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.util.Scanner;

public class Solution {
    public static void main(String[] args) {
	Scanner sc = new Scanner(System.in);
	StringBuilder sb = new StringBuilder();
	int n = sc.nextInt();
	seivePrimes();
	for (int i = 0; i < n; i++) {
	    long a = sc.nextLong();
	    if (isSquaredPrime(a)) {
		sb.append("YES\n");
	    } else {
		sb.append("NO\n");
	    }
	}
	System.out.print(sb.toString());
	sc.close();
    }

    static int maxlim = 1000000;

    static boolean isSquaredPrime(long a) {
	long sqrt = (int) Math.sqrt(a);
	return a >= 4 && sqrt * sqrt == a && isPrime((int) sqrt);
    }

    private static void seivePrimes() {
	for (int i = 3; i * i <= maxlim; i += 2) {
	    if (!isNotPrime[(i - 3) / 2]) {
		for (int j = i; i * j <= maxlim; j += 2) {
		    isNotPrime[(i * j - 3) / 2] = true;
		}
	    }
	}
    }

    static boolean[] isNotPrime = new boolean[(maxlim - 3) / 2 + 1];// isNotPrimes[x]=>2*x+3
								    // is not
    // 778 -> SB
    // 746 -> bit

    private static boolean isPrime(int t) {
	if ((t & 1) == 0)
	    return t == 2;
	else
	    return !isNotPrime[(t - 3) / 2];
    }
}