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;
import java.util.HashSet;
import java.util.Set;
//numbers has exactly three distinct positive divisors=prime*prime
public class Tprimes {
    static int SIZE = 1000001;
    static boolean[] sieve = new boolean[SIZE];

    public static void main(String[] args) throws NumberFormatException,
            IOException {
        computeSieve();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        long n = Long.parseLong(br.readLine().trim());
        String[] s = br.readLine().trim().split("\\s+");
        Set<Long> set = new HashSet<Long>();

        for (int i = 0; i < SIZE; i++) {
            if (sieve[i])
                set.add(((long)i) * i);
        }
        for (int i = 0; i < s.length; i++) {
            long num = Long.parseLong(s[i]);
            if (set.contains(num))
                System.out.println("YES");
            else
                System.out.println("NO");
        }

    }

    public static void computeSieve() {
        for (int i = 0; i < SIZE; i++)
            sieve[i] = true;

        sieve[1] = false;
        for (int i = 2; i < SIZE; i++) {
            if (sieve[i]) {
                // System.out.println(i + " is prime");
                for (int j = i + i; j < SIZE; j = j + i)
                    sieve[j] = false;

            }
        }
    }
}