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Chapter 6 (Part 1) Other Security Overview

Chapter 6 (Part 1) Other Security Overview. Beyond basic cryptography: Secret splitting - divide a message into n pieces, such that a ll n pieces must be combined to recover the message Blind signatures – produce an unlinkable digital signature

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Chapter 6 (Part 1) Other Security Overview

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  1. Chapter 6 (Part 1) Other SecurityOverview • Beyond basic cryptography: • Secret splitting - divide a message into n pieces, such that all n pieces must be combined to recover the message • Blind signatures – produce an unlinkable digital signature • Bit commitment allows a user to commit to a prediction without revealing it • Cryptographic protocols make use of cryptography to accomplish some task securely • Authentication • Key-exchange Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  2. Motivation for Secret Splitting • A professor, Carol, encrypts her grade file with a symmetric-key cryptosystem • Good: only Carol can read grades (privacy) • Good: only Carol can modify grades (integrity) • Bad: if Carol becomes incapacitated nobody else can recover the grades • Carol needs some kind of a mechanism that will allow someone other than her to decrypt the grade file in case of an emergency Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  3. Secret Splitting • Secret splitting makes it possible to divide a message into n piecescalled shadows, such that: • Combining less than n shadows yields nothing • Combining all n shadows yields the message • Carol can split her key into n shadows and give one to n different people she trusts: • Good: if Carol becomes incapacitated all n people can get together and recover the grade file • Good: Unlikely that all n people would betray Carol’s trust Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  4. Secret Splitting Using One-Time Pads • M = “THEKEYISTHREE” • Create four shadows: • Generate n-1 one-time pads (as long as M): • P1 = PDJEUVNSKTUEG • P2 = NBEXUYYKPAQJZ • P3 = ICMKELDAOFGMC • Encrypt M with P1: • C1 = JLOPZUYLEBMJL • T (20) + P (16) mod 26 = J (10) • H (8) + D (4) mod 26 = L (12) • E(5) + J (10) mod 26 = O (15) … Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  5. Secret Splitting Using One-Time Pads (cont) • Encrypt C1 with P2: • C2 = XNTNUTXWUCDTL • Encrypt C2 with P3: • C3 = GQGYZFBXJIKGO • P1, P2, P3, and C3 are the four shadows • Good: all four shadows are required to reconstruct M: • Use P3 to decrypt C3 yielding C2 • Use P2 to decrypt C2 yielding C1 • Use P1 to decrypt C1 yielding M • Bad: What happens if Carol and one of the shadow holders become incapacitated? Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  6. Secret Sharing • Secret sharing (also called a threshold scheme) makes it possible to divide a message into n shadows, such that: • Combining less than k shadows yields nothing • Combining any k (or more) yields the message • Example: • Carol uses a (3-8)-threshold scheme to divide her key into eight shadows (any three required to recover M) • Give three to Alice, one to Bob, two to Dave, and one each to Elvis and Fred • Carol’s key can be recovered by: • Alice (3) • Dave (2) and Bob (1) • Bob (1), Elvis (1), and Fred (1) • Etc. • Good: Need some, not all, shadows to recover the key Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  7. Motivation for Blind Signatures • Dave owns a bank • Carol has an account at Dave’s bank • Carol wants to withdraw some digital money • Carol would like for the digital money to be: • Valid – it should be accepted as payment by merchants (perhaps after some verification procedure) • Anonymous – Dave should not be able to determine where Carol spends her money • Blind signatures allow Dave to create digital money that is both valid and anonymous Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  8. Blind Signatures • Blind signatures enable a user to digitally sign a document without seeing its contents • Assume Dave’s RSA public/private keys are: • Public: e = 413, n = 629 • Private: d = 53 • Before giving a message, m, to Dave for his signature, Carol can blind it • Choose a random blinding factor, b • 1 <b<n-1 • b is relatively prime to n • Example (using small numbers): m = 250 and b = 5 Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  9. Blind Signatures (cont) • Carol blinds the message: B = (m  (be)) mod n B= (250  (5413)) mod 629 B = 172 • Carol gives the blinded message, B, to Dave • Dave cannot read the blinded message because he does not know the blinding factor • If n is large, exhaustive search for b is unfeasible Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  10. Blind Signatures (cont) • Dave signs the blinded message as he would any other message: S’ = Bdmod n S’ = 17253 mod 629 S’ = 168 • Dave sends the signed blinded message, S’, to Carol Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  11. Blind Signatures (cont) • Carol unblinds the signed blinded message by multiplying it by b’s multiplicative inverse modulo n • In our example, b = 5, so b-1 = 126 since (126  5) mod 629 = 1 • Carol computes: S = (S’ b-1) mod n S = (168  126) mod 629 S = 411 • Note that the resulting digital signature, S = 411, is identical to the one that would be produced by Dave signing m = 250 with his private key! Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  12. Properties of Blind Signatures • Validity – as with normal digital signatures: • Anyone can use Dave’s public key to verify his signature on the document is valid • Dave’s signature still cannot be forged or moved to another document, and Dave cannot repudiate his signature • Unlinkability – unlike normal digital signatures: • Dave cannot subsequently link the unblinded signed document to the blind document that he signed Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  13. Unlinkability of Blind Signatures • Suppose: • Carol gives Dave two blinded documents to sign • Dave signs them, returns them to Carol, and keeps copies of the two blind documents • Carol unblinds them and gives Dave copies of the two unblinded documents bearing his signature • Then: • Dave will not be able to determine which unblinded document corresponds to which blinded document Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  14. Example of Unlinkability • Carol gives Dave two blinded documents to sign • B1 = 542, B2 = 492 • Dave signs them, returns them to Carol, and keeps copies of the two blind documents • Carol unblinds them and gives Dave copies of the two unblinded documents bearing his signature • S1 = 217, S2 = 121 • Dave can verify his signature and learn the contents of the documents he signed: • m1 = 217413 mod 629 = 200, m2 = 121413 mod 629 = 100 • Dave cannot link an unblinded document to the corresponding blind document: • B1 = m1 and B2 = m2? • B1 = m2 and B2 = m1? Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  15. Example of Unlinkability (cont) • To link an unblinded document to the corresponding blind document • B1 (542) = m1 (200) and B2 (492) = m2 (100), or • B1 (542) = m2 (100) and B2 (492) = m1 (200) • Dave must determine the blinding factor used to blind each document • Dave can use exhaustive search to find the blinding factors: • b1 = 409 since (100 × 409413) mod 629 = 542 • b2 = 557 since (200 × 557413) mod 629 = 492 • Dave knows that the first blind document he signed, B1, was m2 and the second blind document was m1 • For large values of n, exhaustive search is not feasible and therefore the signatures are unlinkable Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  16. Motivation for Blind Signatures (cont) • Why would Dave sign a blind document that he could not read and create an unlinkable signature? • Recall: • Dave owns a bank • Carol has an account at Dave’s bank • Carol wants to withdraw some digital money • Carol would like for the digital money to be: • Valid – it should be accepted as payment by merchants (perhaps after some verification procedure) • Anonymous – Dave should not be able to determine where Carol spends her money • Blind signatures allow Dave to create digital money that is both valid and anonymous Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  17. Digital Money Without Blind Signatures • Carol creates a message containing a serial number and a value • Serial number = 603482, Value = $10 • Dave signs the message and deducts $10 from Carol’s account • Carol uses the signed message to pay a merchant • The merchant uses Dave’s public key to verify his signature • The merchant redeems the money with Dave for $10 • Good: Carol’s digital money is valid • Bad: Carol’s digital money is not anonymous • Dave could keep a record of each serial number and to whom it was issued • When a merchant redeems digital money Dave could determine to whom that money was issued Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  18. Digital Money With Blind Signatures • Carol creates a message containing a serial number and a value • Serial number = 603482, Value = $10 • Carol blinds the message before giving it to Dave to sign • Dave does not know the blinding factor so he cannot see the contents of the message (e.g., the serial number) • Dave signs the message and deducts $10 from Carol’s account • Carol unblinds the message uses the signed message to pay a merchant • The merchant uses Dave’s public key to verify his signature • The merchant redeems the money with Dave for $10 • Good: Carol’s digital money is valid • Good: Carol’s digital money is anonymous Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  19. Digital Money With Blind Signatures (cont) • Problem #1: double spending • Carol uses her digital money to pay one merchant • The merchant uses Dave’s public key to verify it is valid • Carol uses the same piece of digital money to pay another merchant • The merchant uses Dave’s public key to verify it is valid • Twenty dollars worth of digital money has cost Carol $10 • Solution: merchant must check with Dave and make sure the digital money has not already been spent before accepting it Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  20. Digital Money With Blind Signatures (cont) • Problem #2: fraud • Carol creates a message worth $1000 • Carol blinds the message before giving it to Dave to sign telling him it is worth $10 • Dave does not know the blinding factor so he cannot see the contents of the message • Dave signs the message and deducts $10 from Carol’s account • $1,000 worth of digital money has cost Carol $10 • Solution: Dave needs to be pretty sure of the value of the digital money without actually seeing it (and the serial number) Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  21. Digital Money With Blind Signatures (cont) • Dave requires Carol to create and submit 100 messages: m1 = “Serial number = 935076, Value = $10” m2 = “Serial number = 104766, Value = $10” . . . m100 = “Serial number = 337147, Value = $10” • Carol chooses 100 different blinding factors, b1, b2, …, b100 • Carol uses the blinding factors to create 100 blinded messages • Carol gives all 100 blinded messages to Dave and tells him their value ($10 each in this case) Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  22. Digital Money With Blind Signatures (cont) • Dave chooses 99 of the 100 messages at random to challenge • Dave asks Carol for the corresponding blinding factors • Dave unblinds each of the 99 messages and checks to see that each is worth $10 • If all checks succeed Dave signs the one blind message he did not challenge and returns it to Carol • Carol unblinds the message • Carol now has a valid, anonymous piece of digital money from Dave Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  23. Digital Money With Blind Signatures (cont) • For Carol to get $1,000 worth of digital money for $10 she would have to: • Create 99 messages containing the value $10 • Create one message containing the value $1,000 • Hope that the $1,000 message is the one message that Dave does not challenge • Carol’s chances of succeeding are one in 100 • Dave can lower the odds of fraud by requiring 1,000 or 1,000,000 messages to be submitted Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  24. Motivation for Bit Commitment • Might want to commit to a prediction without revealing it • Chuck and Bill’s virtual coin flips – Scenario #1 • Chuck thinks of a value, either ‘heads’ or ‘tails’ • Bill announces his guess • Chuck tells Bill his value • Problem: Chuck can cheat Bill • Chuck has not committed to a value - he can change it after Bill guesses • Chuck and Bill’s virtual coin flips – Scenario #2 • Bill thinks of his guess • Chuck thinks of a value and announces it to Bill • Bill tells Chuck his guess • Problem: Bill can cheat Chuck • Chuck had to reveal his value in order to commit to it Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  25. Motivation for Bit Commitment (cont) • Chuck and Bill’s virtual coin flips – Scenario #3 • Chuck chooses a value, writes it on a piece of paper, seals it in an envelope, and hands the envelope to Bill • Bill announces his guess • Bill opens the envelope and both learn whether Bill was right • Good: neither can cheat • Fairness requirements: • Chuck must commit to his value in such a way that: • Chuck cannot subsequently change the value • Bill does not learn Chuck’s value until after Bill has guessed Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  26. Bit Commitment • Bit commitment allows someone to commit to a prediction without revealing it • Bit commitment has two phases: • Commitment phase: one party commits to a prediction in such a way that it cannot be subsequently changed • Verification phase: the second party learns the first party’s prediction • Cheating is impossible if: • The prediction cannot be changed after the commitment phase • The prediction is not revealed until the verification phase Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  27. Bit Commitment Using a Symmetric-Key Cryptosystem • Commitment phase: • Chuck chooses a random key, k, and encrypts his prediction • M = Encrypt(p,k) • Chuck gives a copy of M to Bill • Problem: easy for Chuck to change prediction by finding M such that: • Decrypt(M, k1) = 0, and • Decrypt(M, k2) = 1 • Solution: • Bill send a random string of bits, R, to Chuck • Chuck concatenates his prediction to R and then encrypts: • M = Encrypt(“Rp”, k) Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  28. Bit Commitment Using a Symmetric-Key Cryptosystem • Commitment phase: • Bill send a random string of bits, R, to Chuck • Chuck concatenates his prediction to R and then encrypts: • M = Encrypt(“Rp”, k) • Chuck gives a copy of M to Bill • Verification phase: • Chuck sends k to Bill • Bill decrypts M, checks R, and learns p: • Decrypt (M, k) Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  29. Bit Commitment Using a Symmetric-Key Cryptosystem • Neither can cheat: • The prediction cannot be changed after the commitment phase • A good cryptosystem will not allow Chuck to create an M such that: • Decrypt(Rp1,k1) = M • Decrypt(Rp2,k2) = M • The prediction is not revealed until the verification phase • Bill does not know the key Chuck chose and cannot read M without it Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  30. Bit Commitment Using a One-Way Hash Function • Commitment phase: • Chuck creates two random strings of bits, R1 and R2 • Chuck concatenates R1, R2, and his prediction, p, and sends the result through the one-way hash function, H: • h = H(R1R2p) • Chuck sends h and R1 to Bill • Verification phase: • Chuck sends R2 and p to Bill • Bill computes the hash: • h’ = H(R1R2p) • Bill verifies that h’ = h Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  31. Bit Commitment Using a One-Way Hash Function • Neither can cheat: • The prediction cannot be changed after the commitment phase • A good one-way hash function will not allow Chuck to create an R1, R2, and p such that there is a collision: • H(R1R2p1) = h • H(R1R3p2) = h • The prediction is not revealed until the verification phase • Since the hash function is one-way, Bill cannot deduce p from h and R1 Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  32. Advantage of One-Way Hash • Only one message is sent in the commitment phase • As opposed to two messages with symmetric cryptographic approach • Bill sends random string of bits • Chuck appends secret, encrypts and returns it Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  33. Cryptographic Protocols • A protocol is an agreed-upon sequence of actions performed by two or more principals • Cryptographic protocols make use of cryptography to accomplish some task securely • Example: • How can Alice and Bob agree on a session key to protect a conversation? • Answer: use a key-exchange cryptographic protocol Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  34. Key Exchange with Symmetric Cryptography • Assume Alice and Bob each share a key with a Key Distribution Center (KDC) • KA is the key shared by Alice and the KDC • KB is the key shared by Bob and the KDC • To agree on a session key: • Alice contacts the KDC and requests a session key for Bob and her • The KDC generates a random session key, encrypts two copies, one with KA and one with KB, and sends the results to Alice Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  35. Key Exchange with Symmetric Cryptography (cont) • Agreeing on a session key (cont): • Alice decrypts the part of the message encrypted with KA and learns the session key • Alice sends the part of the message encrypted with KB to Bob • Bob receives Alice’s message, decrypts it, and learns the session key • Alice and Bob communicate securely using the session key Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  36. Key Exchange with Symmetric Cryptography (cont) • The key-exchange protocol: A: => KDC (A,B); KDC: => A (E(KAB,KA), E(KAB,KB)); A: => B (E(KAB,KB)); Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  37. Key Exchange with Symmetric Cryptography (cont) • Issues: • Security depends on secrecy of KA and KB • KDC must be secure and trusted by both Alice and Bob • KA and KB should be used sparingly • The use of a new session key for each conversation limits the chances/value of compromising a session key Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  38. Attacking the Protocol • Alice and Bob set up a secure session protected by KAB • An intruder, Mallory, watches them do this and stores the KDC’s message to Alice and all the subsequent messages between Alice and Bob encrypted with KAB • Mallory cryptanalyzes the session between Alice and Bob and eventually recovers KAB • The next time Alice and Bob want to talk Mallory intercepts the KDC’s reply and replays the old message containing KAB • Alice and Bob conduct a “secure” conversation which is protected by KAB which is known to Mallory Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  39. Replay Attack A: => KDC (A,B); KDC: => A (E(KAB,KA), E(KAB,KB)); A: => B (E(KAB,KB)); // Alice and Bob encrypt their messages using KAB // Mallory recovers KAB by analyzing Alice and Bob’s session A: => KDC (A,B); KDC: => A (E(KAB’,KA), E(KAB’,KB)); // Mallory intercepts the above message and replaces it M: => A (E(KAB,KA), E(KAB,KB)); A: => B (E(KAB,KB)); // Mallory reads all traffic session between Alice and Bob Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  40. What Went Wrong? • Alice and Bob need to be able to distinguish between a current (or fresh) response from the KDC and an old one • Solutions: • Alice and Bob could keep track of all previously-used session keys and never accept an old session key • KDC could include freshness information in its messages • Timestamps • Nonces Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  41. Using Timestamps to Establish Freshness A: => KDC (A,B); KDC: => A (E((KAB,TKDC),KA), E((KAB,TKDC),KB)); A: => B (E((KAB,TKDC),KB)); Where TKDC is a timestamp from the KDC’s clock and: • Alice and Bob’s clocks are both synchronized with the KDC’s • Alice and Bob both check the KDC’s message to make sure it was generated recently Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  42. Using Nonces to Establish Freshness • A nonce is a randomly-generated value that: • Is never reused • Can be used to prove the freshness of a message A: => KDC (A,B,NA); B: => KDC (A, B, NB); KDC: => A (E((KAB,NA),KA)); KDC: => B (E((KAB,NB),KB)); Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  43. Key-Exchange with Public-Key Cryptography • Alice learns Bob’s public key (by either asking Bob or some third party) • Alice generates a random session key, KAB • Alice encrypts the session key with Bob’s public key • Alice sends Encrypt(KAB,BPublic) to Bob • Bob receives Alice’s message and decrypts it with his private key • Alice and Bob encrypt their subsequent communications with KAB Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  44. Attacking the Protocol • Recall the man-in-the-middle attack • If Mallory can trick Alice into thinking that MPublic is Bob’s public key • Allice: Encrypt(KAB,MPublic) • Mallory can decrypt Alice’s first message to Bob • Mallory learns the proposed session key KAB • Mallory can send Bob: Encrypt(KAB,BPublic) • Alice and Bob will encrypt their subsequent communications with KAB thinking that it is secure • This is a very serious problem because it’s often difficult to be sure you know somebody’s public key Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  45. The Interlock Protocol • Combating the man-in-the-middle attack: • Alice and Bob exchange public keys • Alice encrypts her message using Bob’s public key. Alice sends half the encrypted message to Bob (e.g., every other bit) • Bob encrypts his message using Alice’s public key. Bob sends half the encrypted message to Alice (e.g., every other bit) • Alice sends the other half of her encrypted message to Bob. Bob puts the two halves together and decrypts them using his private key • Bob sends the other half of his encrypted message to Alice. Alice puts the two halves together and decrypts them using her private key Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  46. The Interlock Protocol (cont) • Foiling the man-in-the-middle: • Assume Mallory can trick Alice into using MPublic instead of BPublic • When Mallory receives the first half of Alice’s message she won’t be able to decrypt it and re-encrypt it with BPublic • Mallory must invent a completely new message, encrypt it and send half of it to Bob • When the second half of Alice’s message arrives, Mallory can put the two halves together, decrypt, and learn what Alice’s original message was • However, Mallory has already committed to the first half of the message and it is too late to change • Therefore, Bob will not get the message Alice sent, and Alice and Bob will probably be able to figure out that there is an intruder between them Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  47. Authentication • Authentication is the process of proving your identity to someone else • One-way • Two-way • Authentication protocols are often designed using a challenge and response mechanism • Authenticator creates a random challenge • Authenticatee proves identity by replying with the appropriate response Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  48. One-way Authentication Using Symmetric-Key Cryptography • Assume that Alice and Bob share a secret symmetric key, KAB • One-way authentication protocol: • Alice creates a nonce, NA, and sends it to Bob as a challenge • Bob encrypts Alice’s nonce with their secret key and returns the result, Encrypt(NA, KAB), to Alice • Alice can decrypt Bob’s response and verify that the result is her nonce A: => B(NA); B: => A(Encrypt(NA, KAB)); • Authenticates Bob Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  49. Two-way Authentication A: => B(NA); B: => A(NB, Encrypt(NA, KAB)); A: => B(Encrypt (NB, KAB)); Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

  50. One-way Authentication Using Symmetric-Key Cryptography • Problem: an adversary, Mallory, might be able to impersonate Bob to Alice: • Alice sends challenge to Bob (intercepted by Mallory) • Mallory does not know KAB and thus cannot create the appropriate response • Mallory may be able to trick Bob (or Alice) into creating the appropriate response for her: A: => M(NA); M: => B(NA); B: => M(Encrypt(NA, KAB)); M: => A(Encrypt(NA, KAB)); Fundamentals of Secure Computer Systems Chapter 6 – Other Security Building Blocks

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