return (pow(2,n+1,1000000007)+2)%1000000007

My normal solution

https://github.com/joy-mollick/Problem-Solving-Solutions-Math-Greedy-/blob/master/HackerRank-Sherlock%20and%20Square.cpp

the trick is to use a custom code for power generation

Is there any problem with the logic?

#include <bits/stdc++.h>
#define MOD 1000000007
using namespace std; 

int main(){
    int t; 
    cin>>t;
    for(int i=0;i<t;i++){
        long n; 
        cin>>n;
        cout<<2*fmod(1+pow(2,n),MOD)<<"\n";
    }
    return 0;
}

can anyone tell me why am I getting termination due to timeout for the following code

test = int(input())

for i in range(test):
    num = int(input())
    answer = (2**(num+1)) + 2
    print(answer % 1000000007)

is there a way to do it even faster