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 Solution230B{

	public static long MAX_I = 1000010;
	public static int[] tPrime = new int[1000010];

	public Solution230B() {


	}	

	public static void makeTprime() {

		tPrime[0] = 1;
		tPrime[1] = 1;

		for (int i = 2 ; i < MAX_I ; i++) {
			
			if(tPrime[i] == 1) 
				continue;

			for (int j = 2*i ; j < MAX_I ; j+=i) {
				tPrime[j] = 1;
			}
		}

	}

	public static void main(String[] args){
		makeTprime();

		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();

		for (int i = 0 ; i < n ; i++) {
			
			long input = sc.nextLong();
			
			long root = (long) Math.sqrt(input);

			if( (root*root != input) || (tPrime[(int)root] == 1) ) {
				System.out.println("NO");
			} else {
				System.out.println("YES");
			}

		}

	}

}