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 Solution{
   public static void main(final String[] args) throws IOException {
       final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
       String input = br.readLine();
       StringTokenizer st = new StringTokenizer(input, " ");

       final int total = Integer.parseInt(st.nextToken());
       input = br.readLine();
       st = new StringTokenizer(input, " ");

       final long[] in = new long[total];

       for (int i = 0; i < in.length; i++) {
           in[i] = Long.parseLong(st.nextToken());
       }

        int size = (int) Math.pow(10, 6);
        boolean[] primes = new boolean[size];
        HashMap<Long, Long> map = new HashMap<Long, Long>();
        
        for(int k = 2; k*k<primes.length; k++){
            if(!primes[k]){
                for(int l = k; l*k<primes.length; l++){
                    primes[l*k] = true;
                }
            }
        }

        for(int j = 2; j<primes.length; j++){
            if(!primes[j]){
                map.put((long) j*j, (long) j*j);
            }
        }

        for(int i = 0; i<in.length; i++){
            if(map.containsKey((long)in[i])){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
            }
        }
    }
}