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.IOException;
import java.io.PrintWriter;
import java.util.Scanner;

public class Test {
    public static void main(String[] args) throws IOException{
        try (   PrintWriter printWriter = new PrintWriter(System.out);
                Scanner scanner = new Scanner(System.in)
                ){
            int max = 1000000;
            boolean seive [] = new boolean [max + 1];
            for (int i = 0; i < max + 1; i++) {
                seive[i] = true;
            }
            for (int i = 2; 2*i < max + 1; i++) {
                if (seive[i])
                    for (int j = 2* i; j < max + 1; j+=i) {
                        seive[j] = false;
                    }
            }
            seive[1] = false;
            int n = scanner.nextInt();
            for (int i = 0; i < n; i++) {
                long temp = scanner.nextLong();
                if ((int)Math.sqrt(temp) == Math.sqrt(temp) && seive[(int)Math.sqrt(temp)]){
                    printWriter.print("YES\n");
                }
                else
                    printWriter.print("NO\n");
            }
        }
    }
    
}