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.Arrays;
import java.util.Scanner;

public class TPrimes {
	static boolean[] prime = new boolean[1000001];
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		prime();
		for(int i = 0; i < n; i++) {
			long temp = scanner.nextLong();
			if(Math.sqrt(temp)==Math.round(Math.sqrt(temp))&&temp!=1) {
				if(prime[(int) Math.sqrt(temp)]) System.out.println("YES");
				else System.out.println("NO");
			}
			else System.out.println("NO");
		}
	}
	public static void prime() {
		Arrays.fill(prime, true);
		for(int j = 2; j < prime.length; j++) {
			for(int i = j*2; i < prime.length; i+=j) prime[i] = false;
		}
	}
}