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 Tprimes {

	public static void main(String[] args) {
		int max = 1000005;
		boolean[] isPrimes = new boolean[max + 1];
		for (int i = 0; i < max; i++) {
			isPrimes[i] = true;
		}
		isPrimes[0] = isPrimes[1] = false;
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		for (int i = 0; i * i < max; i++) {

			if (isPrimes[i] == true)
				for (int j = i * i; j <= max; j += i) {
					isPrimes[j] = false;
				}
		}

		for (int i = 0; i < n; i++) {
			long x = sc.nextLong();
			double sqrt = Math.sqrt(x);
			int intSqrt = (int) sqrt;
			if (Math.ceil(sqrt) != Math.floor(sqrt))
				System.out.println("NO");
			else if (isPrimes[intSqrt])
				System.out.println("YES");
			else
				System.out.println("NO");
		}
	}
}