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.*;
public class Lo
{
	static boolean isPrime(int n)
	{
		if(n<=1)
		{
			return false;
		}
		if(n<=3)
		{
			return true;
		}

		if(n%2==0 || n%3==0)
		{
			return false;
		}

		for(int i=5;i*i<=n;i+=6)
		{
			if(n%i==0 || n%(i+2)==0)
			{
				return false;
			}
		}
		return true;
	}

	static boolean isThree(long n)
	{
		int sq = (int)(Math.sqrt(n));
		if(1L * sq*sq!=n)
		{
			return false;
		}

		return isPrime(sq)?true:false;
	}
     
   
   
     
    public static void main(String[] args) 
    {
    	Scanner s = new Scanner(System.in);
    	int n = s.nextInt();
    	for(int i=0;i<n;i++)
    	{
    		long p =s.nextLong();
    		if(isThree(p))
    		{
    			System.out.println("YES");
    		}
    		else
    		{
    			System.out.println("NO");
    		}
    	}
    }
}