Please help to answer below. I have seen other posts about this questions but none of them are correct. PLEASE DO NOT COPY-PASTE FROM OTHER INCORRECT ANSWERS HERE AGAIN, I WILL DISLIKE. THANKS.

Use the student number as 6961234.

This is the code from lecture notes.

5 Task Five: 3-bit CFB (3 marks) This task comprises three parts. Part one. This task is to implement a 3-bit CFB TEA algorithm. See the code for TEA in the lecture notes. Part two. Encrypt your student number using 3-bit CFB TEA algorithm as devised above. Part three. Add all the digits of your student number mod 7. Let the result be c. Then implement c-bit CFB TEA algorithm, and encrypt your student number using this algorithm. Compare the time needed to encrypt using 3-bit CFB TEA algorithm and c-bit CFB TEA algorithm, and write your result in Task5Report.PDF. Explain what has happened. Remark: If the result of the modulo operation is 0 or 3 , that is c=0 or c=3, then you need to implement 2-bit CFB TEA instead. TEA Encryption void decrypt(unsigned long v, unsigned long k) \{ unsigned long v0=v[0],v1=v[1], sum =0C6EF3720,i;/ setup * unsigned long delta =09e3779b9;/ key schedule constant * unsigned long k0=k[0], k1=k[1], k2=k[2], k3=k[3]; /* cache key */ for (i=0;i>)+k3; v0=(v1>)+k1; sum = delta; / end cycle * \} v[0]=v0;v[1]=v1 \}