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.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        HashSet<Long> prime = new HashSet<Long>();
        HashMap<Long, Boolean> map = new HashMap<Long, Boolean>();
        int size = 1000000 + 1;
        boolean[] arr = new boolean[size];
        Arrays.fill(arr, true);
        arr[0] = arr[1] = false;

        for (int i = 2; i < arr.length; i++) {
            if (arr[i]) {
                prime.add(i * 1L * i);
                map.put((i * 1L * i), true);

                for (int j = i + i; j < arr.length; j += i) {

                    arr[j] = false;

                }

            }
        }

        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        for (int i = 0; i < n; i++) {
            if (map.containsKey(input.nextLong())) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }

    }
}