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.HashSet;
import java.util.Scanner;
import java.util.Vector;

public class B230 {

	static boolean prime[] = new boolean[1000001];
	static HashSet<Integer> set = new HashSet<>();

	public static void seive(int n) {
		prime[0] = false;
		prime[1] = false;

		for (int i = 2; i < 1000001; i++)
			prime[i] = true;

		for (int i = 2; i < 1000001; i++) {
			// If prime[p] is not changed, then it is a prime
			if (prime[i] == true) {
				// Update all multiples of p
				for (int j = 2; i * j < 1000001; j++)
					prime[i * j] = false;
			}
		}
	}

	public static boolean isPrime(Double num) {
		// System.out.println("in isPrime() "+vector.size()+" num is "+num);
		return set.contains(num.intValue());
	}/*
		 * 
		 * static double power(double x, double y, double p) { double res = 1;
		 * // Initialize result x = x % p; // Update x if it is more than or //
		 * equal to p while (y > 0) { // If y is odd, multiply x with result if
		 * (y%2!=0) res = (res*x) % p;
		 * 
		 * // y must be even now y = y/2; // y = y/2 x = (x*x) % p; } return
		 * res; }
		 * 
		 * static // This function is called for all k trials. It returns //
		 * false if n is composite and returns false if n is // probably prime.
		 * // d is an odd number such that d*2<sup>r</sup> = n-1 // for some r
		 * >= 1 boolean miillerTest(double d, double n) { // Pick a random
		 * number in [2..n-2] // Corner cases make sure that n > 4 double a = 2
		 * + Math.random() % (n - 4);
		 * 
		 * // Compute a^d % n double x = power(a, d, n);
		 * 
		 * if (x == 1 || x == n-1) return true;
		 * 
		 * // Keep squaring x while one of the following doesn't // happen //
		 * (i) d does not reach n-1 // (ii) (x^2) % n is not 1 // (iii) (x^2) %
		 * n is not n-1 while (d != n-1) { x = (x * x) % n; d *= 2;
		 * 
		 * if (x == 1) return false; if (x == n-1) return true; }
		 * 
		 * // Return composite return false; }
		 * 
		 * // It returns false if n is composite and returns true if n // is
		 * probably prime. k is an input parameter that determines // accuracy
		 * level. Higher value of k indicates more accuracy. static boolean
		 * isPrime(double n, double k) { // Corner cases if (n <= 1 || n == 4)
		 * return false; if (n <= 3) return true;
		 * 
		 * // Find r such that n = 2^d * r + 1 for some r >= 1 double d = n - 1;
		 * while (d % 2 == 0) d /= 2;
		 * 
		 * // Iterate given nber of 'k' times for (int i = 0; i < k; i++) if
		 * (miillerTest(d, n) == false) return false;
		 * 
		 * return true; }
		 */

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		seive(1000001);
		Scanner ui = new Scanner(System.in);
		int num = ui.nextInt();
		int i = 0;
		long sqrt;
		long current;
		long array[] = new long[num];
		while (i < num) {
			array[i] = ui.nextLong();
			i++;
		}
		i = 0;
		while (i < num) {
			sqrt = (long) Math.sqrt(array[i]);
			if(sqrt-Math.sqrt(array[i])==0)
			{
				if(prime[(int) sqrt]==true)
				{
					System.out.println("YES");
				}
				else
				{
					System.out.println("NO");
				}
			}
			else
			{
				System.out.println("NO");
			}
			i++;
		}
	}

}