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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		long long int num;
		cin >> num;
		double num1 = num * log10( double (num));
		long long int num2 = (long long int) num1;
		double num3 = num1 - num2;
		num = (long long int) pow(10,num3);
		cout << num<< endl;
	}
	return 0;
}