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;
import java.lang.Math;

public class Main {
    private static Boolean nod(long n) {
        long s = Math.round(Math.sqrt(n));
        if (s * s != n || s == n) {
            return false;
        }

        Double ss = Math.sqrt(s);

        for (long i = 2; i <= ss; i++) {
            if (s % i == 0) {
                return false;
            }
        }

        return true;
    }

    private static Boolean check(long s) {
        Double ss = Math.sqrt(s);

        for (long i = 2; i <= ss; i++) {
            if (s % i == 0) {
                return false;
            }
        }

        return true;
    }

    public static void main(String[] args) {
        byte[] a = new byte[1000001];
        for (int i = 2; i <= 1000000; i++) {
            for (int j = 2 * i; j <= 1000000; j += i) {
                a[j] = 1;
            }
        }

        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        for (int i = 0; i < n; i++) {
            long x = scanner.nextLong();
            double s = Math.sqrt(x);
            if (x > 1 && s == (int)s && a[(int)s] == 0) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
    }
}