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:
//package solution;

import java.util.Scanner;

public class Solution {
	public static int MAX = 1000001;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long x;
		
		boolean[] isPrime = new boolean[MAX];
		for(int i = 0; i < MAX; ++i) isPrime[i] = true;
		
		isPrime[1] = false;
		for(int i = 2; i < MAX; ++i) {
			if(isPrime[i]) {
				for(int j = 2; j * i < MAX; ++j) isPrime[i * j] = false;
			}
		}
		
		for(int i = 0; i < n; ++i) {
			x = sc.nextLong();
			long x1 = (long) Math.sqrt(x);
			
			if(x1 * x1 == x && isPrime[(int) x1]) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}
		}
	}
}