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.
 

/*求N^N的最低位。只能观察尾数的规律。 
很容易发现0,1,5,6的任何次幂都为本身。
2,3,7,8是每四次幂循环一次。4,9均是每两次幂一循环 
*/
#include<stdio.h>
int a[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};
int main()
{
	int test,n;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%d",&n);
		if(n%10==0||n%10==1||n%10==5||n%10==6)
		printf("%d\n",a[n%10][0]);
		if(n%10==4||n%10==9)
		printf("%d\n",a[n%10][n%2]);
	    if(n%10==2||n%10==3||n%10==7||n%10==8)
	    printf("%d\n",a[n%10][n%4]);
	}
	return 0;
}