the logic seems fine.
Lets play game ...... m getting 2 WA
@imavneet
Your code for sieve is fine
The solution for the problem is just printing 2^(P-1)
Where P is number of prime numbers till N
See this for further detail
@Aarnav-Jindal-1059677350830863 this code is a bit difficult to understand … could you tell me what difference would it make if i start with 2 and subtract 1 from k later?
could you please tell where the error is … so that i can correct my code
@imavneet
This is what I wanted to show you in that post
Our objective is to factor n! into p and q such that hcf(p,q)=1 and p<q. so we calculate number of distinct primes in n!
let prime factorisation of n!= (p1)^x1 * (p2)^x2 … (pk)^xk
now for each i, (pi)^xi can either go in p or q. Hence 2^k ways. but as we want p<q. therefore only 2^(k-1) ways. That’s why in code, counting primes is started from 3 because ultimately we would have to subtract 1 from k.
Your code is doing a lot of unnecessary computation. No benefit trying to solve a question wrong way buddy. Better learn the correct approach and move ahead no 
If there is anything else you need help me please let me know else please close the doubt
