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 Source {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		boolean[] p = sieveOfEratosthenes((int) Math.pow(10, 6));

		long n = scan.nextLong();
		for (long i = 0; i < n; i++) {
			long x = scan.nextLong();
			long root = (long) Math.sqrt(x);
			if (p[(int) root] && (root * root) == x && x != 1) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}
		}
	}

	public static boolean[] sieveOfEratosthenes(int n) {
		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i < n; i++)
			prime[i] = true;

		for (int p = 2; p * p <= n; p++) {
			if (prime[p] == true) {
				for (int i = p * 2; i <= n; i += p)
					prime[i] = false;
			}
		}

		return prime;
	}
}