2
3
4

2
2
Hint
In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2.In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.
#include<cstdio>
#include<cstring>
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            __int64 n;
            cin>>n;
            double x=n*log10(n*1.0); 
            x-=(__int64)x;
            int a=pow(10.0, x);//a为10的小数次方的整数部分 
            cout<<a<<endl;
        }
    }
    return 0; 
}