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.Scanner;

public class Solution {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		

		boolean[] prime = new boolean[1000000+1];
		for(int i=0; i < prime.length; i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <= prime.length-1; p++){ 
            if(prime[p]) { 
                for(int i = p*2; i <= prime.length-1; i += p) 
                    prime[i] = false; 
            }
		} 
		
		int n = sc.nextInt();
		while(n-- > 0){
			long a = sc.nextLong();
			long sqrt = (long) Math.sqrt(a);
			System.out.println(a==1 ? "NO" : sqrt*sqrt == a && prime[(int) sqrt] ? "YES" : "NO");
		}

		sc.close();
	}
}