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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

        int n = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        //para no hacer i - 1
        int max = 1000001;
        boolean[] noPrime = new boolean[max];
        calculatePrimes(noPrime);

        for(int i = 0; i < n; i++) {
            long x = Long.parseLong(st.nextToken());
            double sqrt = Math.sqrt(x);
            boolean prime = false;
            if (sqrt == (long)sqrt) {
                prime = !noPrime[(int)sqrt];
            }
            bw.write(prime ? "YES" : "NO");
            bw.newLine();
        }
        bw.flush();
    }

    private static void calculatePrimes(boolean[] noPrime) {
        noPrime[0] = true;
        noPrime[1] = true;
        for (int i = 2; i < noPrime.length; i++) {
            if (!noPrime[i]) {
                //error if i * i = int number
                //long because 10^12 > 2^31 2.000.000.000
                for (long j = (long)i * i; j < noPrime.length; j += i) {
                    noPrime[(int)j] = true;
                }
            }
        }
    }
}