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.io.*;
import java.util.*;

public class Main{

	public static boolean[] prime = new boolean[1000001];

	public static void main(String[] args) throws IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(br.readLine());

		String[] input = br.readLine().split(" ");

		StringBuffer sb = new StringBuffer();

		getPrime();

		for(int i = 0; i < T; i++) {
			long curr = Long.parseLong(input[i]);

			sb.append(isTPrime(curr) + "\n");
		}

		System.out.println(sb.toString());

	}

	public static String isTPrime(long n) {
		//1. check if its sqrt is integer
		double sqrt = Math.sqrt(n); 
		if(sqrt != (int)sqrt) {
			return "NO";
		}
		else {
			return isPrime((int)sqrt) ? "YES" : "NO";
		}
	}

	public static boolean isPrime(int n) {
		return prime[n];
	}

	public static void getPrime() {
		Arrays.fill(prime, true);

		prime[1] = false;
		
		for(int i = 2; i <= 1000000; i++) {
			if(!prime[i]) continue;

			for(int j = 2; j * i <= 1000000; j++) {
				prime[i * j] = false;
			}
		}
	}

}