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.BufferedOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.util.BitSet;
import java.util.Scanner;

public class Tprimes {
    public static void main(String[] args) throws IOException {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        long[] nums = new long[n];
        long largest = -1;
        for (int i = 0; i < n; i++) {
            nums[i] = scanner.nextLong();
            if (nums[i] > largest) {
                largest = nums[i];
            }
        }
        scanner.close();

        largest = (long) Math.sqrt(largest);
        if (largest > Integer.MAX_VALUE) {
            largest = Integer.MAX_VALUE;
        }
        int[] primes = primes((int) largest);

        BitSet out = new BitSet(n);

        long sqrt;
        long num;
        boolean good;
        for (int i = 0; i < n; i++) {
            num = nums[i];
            sqrt = (long) Math.sqrt(num);
            good = false;
            if (sqrt * sqrt == num) {
                good = true;
                if (num == 1) {
                    good = false;
                }
                for (int p : primes) {
                    if (p * p > sqrt) {
                        break;
                    }
                    if (sqrt % p == 0 && sqrt != p) {
                        good = false;
                        break;
                    }
                }
            }
            out.set(i, good);
        }
        OutputStream spew = new BufferedOutputStream(System.out);
        for (int i = 0; i < n; i++) {
            if (out.get(i)) {
                spew.write(("YES" + "\n").getBytes());
            } else {
                spew.write(("NO" + "\n").getBytes());
            }
        }
        spew.flush();
    }

    private static int[] primes(int x) {
        x = x + 2;
        BitSet prime = new BitSet(x);
        prime.flip(2, x);
        for (int p = 2; p <= x; p++) {
            if (prime.get(p)) {
                for (int i = p * 2; i <= x; i += p) {
                    if (i < 0)
                        break;
                    prime.set(i, false);
                }
            }
        }
        int[] primes = new int[prime.cardinality()];
        for (int i = 2, j = 0; i < prime.length(); i++) {
            if (prime.get(i)) {
                primes[j] = i;
                j++;
            }
        }
        return primes;
    }
}