Problem: We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer t Т-prime, if t has exactly three distinct positive divisors.

You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.

Input

The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains n space-separated integers xi (1 ≤ xi ≤ 1012).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.

Output

Print n lines: the i-th line should contain "YES" (without the quotes), if number xi is Т-prime, and "NO" (without the quotes), if it isn't.

Examples

Input

3
4 5 6


Output

YES
NO
NO

Note

The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.*;

public class CF
{
	public static void main (String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		int num = 1000000;
		boolean[] isPrime = new boolean[num+1];
		for(int i=2; i<=num; i++)
		    isPrime[i] = true;
		for(int i=2; i*i<=num; i++)
		{
		    if(isPrime[i])
		    {
		        for(int j=i; i*j<=num; j++)
		        {
		            isPrime[i*j]=false;
		        }
		    }
		}
		while(T-->0)
		{
		    long n = sc.nextLong();
		    long sq = (long)Math.sqrt(n);
		    if((sq*sq == n) && isPrime[(int)sq])
		        System.out.println("YES");
		    else
		        System.out.println("NO");
		}
	}
}