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.*;

public class Solution {
    public static void main(String[] args) {
        boolean[] p = new boolean[1000001];
        Arrays.fill(p, true);
        p[0] = p[1] = false;
        for (int i = 2; i <= 1001; ++i) {
            if (p[i]) {
                for (int j = i * i; j < p.length; j += i) {
                    p[j] = false;
                }
            }
        }

        HashSet<Long> tp = new HashSet<>();
        for (int i = 0; i < p.length; ++i) {
            if (p[i]) {
                long v = i;
                tp.add(v * v);
            }
        }

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i = 0; i < n; ++i) {
            long x = sc.nextLong();
            System.out.println(tp.contains(x) ? "YES" : "NO");
        }
    }
}