Alice, a student of grade 66, is thinking about an Olympian Math problem, but she feels so despair that she cries. And her classmate, Bob, has no idea about the problem. Thus he wants you to help him. The problem is:

We denote k!:

k! =1×2×⋯×(k−1)×k

We denote S:

S =1×1!+2×2!+⋯+ (n−1)×(n−1)!

Then S module n is ____________

You are given an integer nn.

You have to calculate S modulo n.

Input
The first line contains an integer T(T≤1000), denoting the number of test cases.

For each test case, there is a line which has an integer n.

It is guaranteed that 10^{18}2≤n≤10 ^18
 .

Output
For each test case, print an integer SS modulo nn.

Hint
The first test is: S = 1\times 1!= 1S=1×1!=1, and 1 modulo 2 is 1.

The second test is: S = 1\times 1!+2 \times 2!= 5S=1×1!+2×2!=5 , and 5 modulo 3 is 2.

样例输入 
2
2
3
样例输出 
1
2

题意：给出n，S=1+2*2!+3*3!+...(n-1)*(n-1)!     求S%n

思路：输出n-1

推导来自：https://zhidao.baidu.com/question/364445479.html

1x1!+2x2!+3x3!+4x4!+.....+nxn!=(n+1)!-1
证明：
左边=(2-1)x1!+2x2!+3x3!+4x4!+.....+nxn!
=2!-1+2x2!+3x3!+4x4!+.....+nxn!
=(1+2)x2!-1+3x3!+4x4!+.....+nxn!
=3!-1+3x3!+4x4!+.....+nxn!
=(1+3)x3!-1+4x4!+.....+nxn!
=4!-1+5x5!+.....+nxn!
..........
=(n+1)!-1
所以：S=1×1!+2×2!+3×3!+……+n×n!=(n+1)!-1

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN=1e5+5;

ll n;

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