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Note: ii k is reused for different messages rn,one could solve for x so it is not secure.
If k is reused for the same m, we obtain thesame signature so this is not a problem. Jf k isdilferent for the same m, it should be randomand unknown (any knov relation betseen the
two k-s allows to solve for 4)
Sottomhne: All of the above are enforced by k =chosen at random from Z.qM for large enough q
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