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.io.*;
import java.util.*;

public class Main {
  static int n;
  static boolean p[] = new boolean[1_000_000 + 16];

  public static void main(String[] args) throws Exception {
    Scanner in = new Scanner(System.in);
    PrintWriter out = new PrintWriter(System.out);

    n = in.nextInt();
    sieve();

    while (n -- > 0) {
      long x = in.nextLong();
      int s = (int) (Math.sqrt(x) + 0.5);

      out.println(((long)s * s == x && !p[s]) ? "YES" : "NO");
    }

    out.flush();
  }

  public static void sieve() {
    p[0] = p[1] = true;
    for (int i = 2; i * i <= 1_000_000; ++i)
      if (! p[i])
        for (int j = i * i; j <= 1_000_000; j += i)
          p[j] = true;
  }
}