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;

public class Cff { // 🙂🙂
    public static void main(String []a) {
        Scanner reader = new Scanner(System.in);

        int t = reader.nextInt();
        boolean []prime = getPrimes(1000000);

        long n;
        long sqrtN;
        while (t-- > 0) {
            n = reader.nextLong();
            sqrtN = (long)Math.sqrt(n);
            System.out.println(
                    (prime[(int)sqrtN] && sqrtN*sqrtN == n && n > 1? "YES": "NO")
            );
        }
    }

    static boolean []getPrimes(int limit) {
        boolean []prime = new boolean[1000001];


        for (int i = 2; i <= limit; i++) {
            prime[i] = true;
        }

        for (int i = 2; i <= 10000; i++) {
            if (isPrime(i)) {
                prime[i] = true;
                for (int j = 2*i; j <= limit; j += i) {
                    prime[j] = false;
                }
            }
        }

        return prime;
    }

    static boolean isPrime(long n) {

        for (long i = 2; i*i <= n; i++) {
            if (n % i == 0) {
                return false;
            }
        }

        return n > 1;
    }
}