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.

樣例輸入 
2
2
3
樣例輸出 
1
2

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

思路：輸出n-1

1x1!+2x2!+3x3!+4x4!+.....+nxn!=(n+1)!-1
證明：
左邊=(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

程式碼：

#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;
}