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.nextInt();
public class Solution {
    public static void main(String[] args) {
        boolean[] res = CalcularPrimos(1000000);
        
        Scanner scanner = new Scanner(System.in);
        int test_case = scanner.nextInt();
            
            for(int z = 0; z < test_case; ++z)
            {               
                long t = scanner.nextLong();
                long root = (long)Math.sqrt(t);    
                if(t == root) System.out.println("NO");
                else if(root*root == t)
                {
                    if(!res[(int)root]) System.out.println("YES");
                    else System.out.println("NO");
                }
                else System.out.println("NO");
            }
                    
    }
    
    public static boolean[] CalcularPrimos(long n)
    {
        boolean[] res = null;
        if(n < 2) return null;
        else
        {
            res = new boolean[(int)(n+1L)];
            for(int i = 2; i <= n; ++i)
            {
                if(!res[i])
                {
                    for(int j = i+i; j <= n; ++j)
                    {
                        res[j] = true;
                        j += i-1;
                    }
                }
            }
            return res;
                   
        }
        
    }
}