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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Created by constanzafierro on 10-06-16.
 */
public class Problem230B {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String[] line = br.readLine().split(" ");
        long[] numbers = new long[n];
        long max = Long.MIN_VALUE;
        for(int i = 0; i < n; i++) {
            numbers[i] = Long.parseLong(line[i]);
            max = Math.max(max, numbers[i]);
        }
        int t = 1000001;
        int[] criba = new int[t];
        for(int i=0; i<t; i++){
            criba[i] = i;
        }
        for(int i=2; i*i<t; i++){
            if(criba[i] == i){
                for(int j = 2; i*j<t; j++){
                    criba[i*j] = i;
                }
            }
        }
        for(long num: numbers) {
            double sqrt_num = Math.sqrt(num);
            if(sqrt_num%1 != 0 || sqrt_num==1) {
                System.out.println("NO");
            }
            else {
                if(criba[(int)sqrt_num] == sqrt_num ) System.out.println("YES");
                else System.out.println("NO");
            }
        }
    }
}