Question: When running the Diffie-Hellman protocol with Anna, Tom accidentally uses q rather than p as the modulus for all the computation. What would be the

When running the Diffie-Hellman protocol with Anna, Tom accidentally uses q rather than p as the modulus for all the computation. What would be the result?

1: Anna and Tom can still obtain the same key at the end of the protocol because q divides p 1

2: The adversary can compute the key calculated by Anna

3: Anna and Tom cannot obtain the same key at the end of the protocol

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