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 tprimes {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int i,j ;
    boolean[] seive = new boolean[1000001];
    Arrays.fill(seive,true);
    seive[0] = seive[1] = false ;
    for(i=2;i<=1000;i++) {
      if(seive[i]) {
        for(j=i*i;j<=1000000;j+= i) {
          seive[j] = false ;
        }
      }
    }
    int r = Integer.valueOf(sc.nextLine()) ;
    String inp = sc.nextLine();
    String[] a = inp.split(" ");
    for(int y=0;y<r;y++) {
      long number = Long.valueOf(a[y]);
      double aroot = Math.sqrt(number);
      if((int)aroot != Math.ceil(aroot)) {
        System.out.println("NO") ;
      }
      else {
        if(seive[(int)aroot]) {
          System.out.println("YES");
        }
        else {
          System.out.println("NO");
        }
      }
    }
  }
}