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;

/**
 *
 * @author tortita
 */
public class Prueba {

    static public int[] llenarCriba(){
        int N = 2000000;
        int[] criba = new int[N];
        criba[0] = 2;
        criba[1] = 2;
        for (int i = 2; i < N; i++) {
            if (criba[i] == 2 || criba[i] == 1){
                continue;
            }
            else{
                criba[i]=1;
                for (long j = i; j*i< N; j++) {
                    criba[(int)j*i]=2;
                }
            }
        }
        return criba;
    }
    
    static public void isTPrime(long n, int[] criba){
        if(n > 4 && n%2 == 0){
            System.out.println("NO");
        }
        else{
            double sqrt = Math.sqrt(n);
            int isqrt = (int) sqrt;
            if ((double) isqrt - sqrt == 0) {
                if (criba[(int) sqrt] == 1) {
                    System.out.println("YES");
                } else {
                    System.out.println("NO");
                }
            } else {
                System.out.println("NO");
            }
        }
    }
    
    static public void main(String[] args){
        int[] criba = llenarCriba();
        Scanner entrada = new Scanner(System.in); // Con esto creamos un objeto Scanner
        int nIntegers = entrada.nextInt();
        for (int i = 0; i < nIntegers; i++) {
            long number = entrada.nextLong();
            isTPrime(number,criba);
        }       
    }
}