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:
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.util.Scanner;



public class Main {

    public static void main(String[] args) {
        int[] a = new int[1000001];
        for(int i=2;i*i<=1000000;i++)
        {
            if(a[i] == 0)
            {
                for(int j=i*i;j<a.length;j+=i)
                {
                    a[j] = 1;
                }
            }
        }
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        while(n!=0)
        {
            long q = s.nextLong();
            int x1 = (int) Math.sqrt(q);
            long aux;
            double sq = Math.sqrt((double)q);
            if((sq - Math.floor(sq)) == 0){
                aux=1;
            }else{
                aux=0;
            }
            if(q == 1)
                System.out.println("NO");
            else if(aux==1 && (a[x1] == 0 ))
                System.out.println("YES");
            else
                System.out.println("NO");
 
            n=n-1;
        }
    }
    
}