The problem says that we have a number n. The possible generated numbers from n are 0 to 2^N - 1. Lets say this value is x.

power set of x will generate 2 ^ x sets.

xor is a modulo 2 sum i.e. sum all the elements and take its xor. if the sum is even then it will generate 0 otherwise 1.

from the 2^x sets, half will generate even sum and half will generate odd.

There are 2^(2^n) subsets of S, and a moment's thought will convince you that the xor values are uniformly distributed across the 2^n possible values. So the answer we're looking for is 2^((2^n)-n)%M. Calculating that naively (even with pow) will get you a TLE because the exponent gets excessive for large n. But, since M is prime, we can use Fermat's Little Theorem to reduce the modulus of the exponent mod M-1and wind up with values that pow can deal with in the time limit.

M = 10**9+7
def solve(n):
    exp = pow(2,n,M-1)
    return pow(2,exp-n,M)

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