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.lang.Math;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    final int SIEVE = (int) 1e6 + 1;
    boolean[] isPrime = new boolean[SIEVE];
    Arrays.fill(isPrime, true);
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i <= (int) Math.sqrt(SIEVE); ++i) {
      if (isPrime[i]) {
        for (int j = i; i * j < SIEVE; ++j) {
          isPrime[i * j] = false;
        }
      }
    }
    int n = in.nextInt();
    for (; n > 0; --n) {
      long x = in.nextLong();
      long t = (long) Math.sqrt(x);
      if (t * t != x) {
        System.out.println("NO");
        continue;
      }
      System.out.println(isPrime[(int) t] ? "YES" : "NO");
    }
  }
}