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.*;
import java.io.*;
import java.lang.*;
import java.math.*;
 
public class Tprime{
	public static void main(String[] args) throws IOException{
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int a = Integer.parseInt(in.readLine());
		String[] input = in.readLine().split(" ");
		boolean[] prime = new boolean[10000005];
		for(int i=2;i<=1000;i++){
			for (int j=i*i;j<=1000000;j+=i){
				prime[j] = true;
			}
		}
		for(int i=0;i<a;++i){
			double n = Double.parseDouble(input[i]);
			boolean k = true;
			if(n == 1){
				System.out.println("NO");
				continue;
			}
			double p = Math.sqrt(n);
			if((int) p == p && !prime[(int) p])							
				System.out.println("YES");
			else
				System.out.println("NO");	
		}	
	}
}