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.*;

public class Main{
	static boolean flag[];
	public static void main(String[] args) {
		Scanner sc= new Scanner(System.in);
		Generate();
		int n=sc.nextInt();
		
		
		while(n-->0) {
			long m=sc.nextLong();
			if(Math.sqrt(m)%(int)Math.sqrt(m)==0) {
				if(flag[(int) Math.sqrt(m)]) {
					System.out.println("YES");
				}else {
					System.out.println("NO");
				}
			}else {
				System.out.println("NO");
			}
		}
		
	}
	
	public static void Generate() {
		flag=new boolean[1000001];
		Arrays.fill(flag, true);
		flag[1]=false;
        for (int p = 2; p * p <= 1000000; p++) 
        {
            // If prime[p] is not changed, then it is a
            // prime
            if (flag[p] == true) 
            {
                // Update all multiples of p
                for (int i = p * p; i <= 1000000; i += p)
                    flag[i] = false;
            }
        }
	}
	
}