#include<stdio.h>
int main(){
	int n,m;
	scanf("%d",&n);
	while(n--){
		scanf("%d",&m);
	    long long ans=1,a=2;
	    while(m!=0){
		    if(m%2!=0)ans=ans*a%1000000;
		    a=(a*a)%1000000;  m/=2;
        }
        printf("%lld\n",ans-1);
    }
}//依旧快速幂

#include <stdio.h> // 快速幂取模函数 //B站大佬有专门教汉诺塔模型的 long long hannuota(long long b, long long e, long long m) { long long r = 1; while (e > 0) { if (e % 2 == 1) { r = (r * b) % m; } b = (b * b) % m; e /= 2; } return r; }

int main() { int N; scanf("%d", &N); while (N--) { int m; scanf("%d", &m); // 计算2^m - 1的最后六位，并输出 long long steps = (hannuota(2, m, 1000000) - 1 ) % 1000000; printf("%lld\n", steps); } return 0; }

#include<stdio.h>
int n;
long long m;
long long ksm(long long m)
{
long long sum = 2;
long long p = 1;
while ((p * 2)< m)
{
sum = (sum * sum) % 1000000;
p *= 2;
}
if (p != m)
{
m = m - p;
sum = (ksm(m) * sum) % 1000000;
}
return sum;
}

int main()
{
scanf("%d", &n);
while (n --)
{
scanf("%lld", &m);
printf("%lld\n", ksm (m) - 1);
m = 0;
}
return 0;
}

#include<iostream>
#include<vector>
#include<algorithm>
#include <bits/stdc++.h>
using namespace std;

//投机取巧，找规律，数量大于62500时，每62500个数作为一个循环，保留的六位数为109376
int main() {
    int n;
    cin >> n;
    while (n--) {
        int x;
        cin >> x;
        int sum=1;
        if(x<62500) {
            while (x--) {
                sum *= 2;
                if (sum > 999999)
                    sum %= 1000000;
            }
            cout << sum - 1 << endl;
        }
        else
        {
            x%=62500;
            sum=109376;
            while (x--) {
                sum *= 2;
                if (sum > 999999)
                    sum %= 1000000;
            }
            cout << sum - 1 << endl;
        }
    }
    return 0;
}

薯饼海特供（快速幂版）
#include<stdio.h>
int ksm(int n);
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		int m;
		scanf("%d",&m);
		long long sum=ksm(m);
		printf("%lld\n",sum-1);
	}
}

int ksm(int n)
{
	long long ans=1;
	long long a=2;
	while(n!=0)
	{
		if(n%2!=0)
		{
			ans=ans*a%1000000;
		}
		a=(a*a)%1000000;
		n/=2;
	}
	return ans;
}```