Given a positive integer N, you should output the most right digit of N^N.

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

For each test case, you should output the rightmost digit of N^N.

2
3
4

7
6

Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
 

代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main ()
{
    int n,t,m,i,s;
    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        m=n%10;
        n=n%4;
        s=m*m*m*m;
        s=s%10;
        for (i=0; i<n; i++)
            s*=m;
        s=s%10;
        cout<<s<<endl;
    }
    return 0;
}
/*
1: 1                4: 6,4,6        7: 9,3,1,7,9
2: 4,8,6,2,4        5: 5            8: 4,2,6,8,4
3: 9,7,1,3,9        6: 6,           9: 1,9,1
*/