Problem Description：

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

Input：

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).  

Output：

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

Sample Input：

2

3

4

Sample Output：

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.

程序代码：

#include<bits/stdc++.h>
using namespace std;
int quickmul(int n)
{
  int ans=1;
  int temp=n%10;
  while(n)
  {
    if(n&1)//等价于n%2==1 
      ans=(ans*temp)%10;
    temp=(temp*temp)%10;
    n>>=1;//等价于n=n/2 
  }
  return ans;
}
int main()
{
  int a,n,count;
  cin>>n;
  while(n--)
  {
    cin>>a;
    cout<<quickmul(a)<<endl;
  }
  return 0;
}