#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define N 500

long long Pow(long long a, long long b, long long c)
{
    long long ans = 1;
    a %= c;
    while(b)
    {
        if(b % 2 == 1)
            ans = (ans * a) % c;
        a = (a * a) % c;
        b /= 2;
    }
    return ans;
}//快数幂，求a的b次幂，c表示对c取余

int main()
{
    int t;
    long long n, num;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld", &n);
        num = Pow(2, n - 1, 1000000007);
        printf("%lld\n", num - 1);
    }
    return 0;
}