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.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        String[] nums = in.readLine().split(" ");
        Set<Long> primes = calcPrimeNumbers(1000000);
        for (int i = 0; i < n; i++) {
            if (primes.contains(Long.parseLong(nums[i]))) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
    }

    public static Set<Long> calcPrimeNumbers(int n) {
        int[] lp = new int[n + 1];
        List<Integer> pr = new ArrayList<>();
        pr.add(2);
        for (int i = 3; i < n; i += 2) {
            if (lp[i] == 0) {
                lp[i] = i;
                pr.add(i);
            }
            for (int j = 0; j < pr.size() && pr.get(j) <= lp[i] && i * pr.get(j) <= n; ++j)
                lp[i * pr.get(j)] = pr.get(j);
        }
        Set<Long> set = new HashSet<>();
        for (Integer integer : pr) {
            set.add((long) integer * integer);
        }
        return set;
    }
}