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.util.HashSet;

public class Main {
  public static void main(String[] args) {
    Scanner s = new Scanner(System.in);
    int n = s.nextInt();
    HashSet<Long> primes = new HashSet<Long>();
    int size = 1000000 + 1;
    boolean[] isPrime = new boolean[size];
    isPrime[2] = true; primes.add((long)2*2);
    for (int j = 3; j < size; j += 2) isPrime[j] = true;
    long start = 3;
    while (start < size) {
     if (isPrime[(int)start]) primes.add(start*start);
     for (long j = start*start; j < size; j += start) isPrime[(int)j] = false;
     start++;
    }
    for (int j = 0; j < n; ++j) 
      if (primes.contains(s.nextLong())) {
         System.out.println("YES");
      } else System.out.println("NO");
  }
}