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.math.BigInteger;
import java.util.*;
import java.io.*;
import java.text.*;

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 32768);
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer t;

    static String sn() {
        while (t == null || !t.hasMoreTokens()) {
            try {
                t = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }
        return t.nextToken();
    }

    static int ni() {
        return Integer.parseInt(sn());
    }

    static long nlo() {
        return Long.parseLong(sn());
    }

    public static void main(String args[]) {
        int t = 1;
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    static long mod = (long) 1e9 + 7;

    static long pro(long a, long b) {
        return (a % mod * b % mod) % mod;
    }

    static long add(long a, long b) {
        return (a + b) % mod;
    }

    static long sub(long a, long b) {
        return ((a % mod - b % mod) + mod) % mod;
    }

    static int maxn = 1000005;

    static int pow(int n) {
        return (int) (Math.log10(n) / Math.log10(2));
    }

    static long modpow(long x, long y) {
        long res = 1l;
        x %= mod;
        while (y > 0) {
            if (y % 2 != 0)
                res = pro(res, x);
            x = pro(x, x);
            y /= 2;
        }
        return res;
    }

    static class pair {
        String a;
        int b;

        pair(String x, int y) {
            a = x;
            b = y;
        }
    }

    static long modInverse(long n) {
        BigInteger n1 = new BigInteger(Long.toString(n));
        String s = (n1.modInverse(new BigInteger(Long.toString(mod)))).toString();
        return Long.parseLong(s);
    }

    static boolean v[] = new boolean[1000005];

    static void seive() {
        v[1] = true;
        for (int i = 2; i < 1000005; i++) {
            if (!v[i]) {
                for (int j = 2 * i; j < 1000005; j += i)
                    v[j] = true;
            }
        }
    }

    static boolean primecheck(long n) {
        if (!v[(int) n])
            return true;
        return false;
    }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static void solve() {
        int n = ni();
        seive();
        for (int i = 0; i < n; i++) {
            long a = nlo();
            long h = (long) Math.sqrt(a);
            if (h * h == a && primecheck(h))
                out.println("YES");
            else
                out.println("NO");
        }
    }
}