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.Map;
import java.util.Scanner;
import java.util.regex.Matcher;

public class CF230B {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        long[] arr = new long[n];
        for(int i = 0 ; i < n;i++){
            arr[i] = s.nextLong();
        }
        int[] num = new int[1000001]; // by defualt, an integer is assign value of 0
        for (int i =2; i < 1000001; i++){
            num[i] =0;
        }
        num[0] = 1;
        num[1] = 1;
        // Check if the square is prime number
        for(int i = 2 ; i < 1000001/2;i++){
            if(num[i]==0){
                for(int j = 2;j*i<1000001;j++){
                    if(num[j*i] == 0){
                        num[j*i] = 1;
                    }
                }
//                for (int j = 2; )
            }
        }

        for(int i = 0;i<n;i++){
///            System.out.println((int) Math.sqrt(arr[i]));
            if(Math.sqrt(arr[i])%1==0){
               if(num[(int) Math.sqrt(arr[i])] == 0){
                    System.out.println("YES");
                    continue;
                }
            }
            System.out.println("NO");
        }

    }

}