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 CodeChef
{
	static int N=1000001;
	static boolean isPrime[];
	static void sieve()
	{
		isPrime = new boolean[N];
		Arrays.fill(isPrime, true);
		isPrime[1]=false;
		isPrime[0]=false;
		for(int i=2; i<=Math.sqrt(N); i++)
		{
			if(isPrime[i])
			{
				int j=i+i;
				while(j<N)
				{
					isPrime[j]=false;
					j +=i;
				}
			}
		}
	}
	public static void main(String[] args) throws IOException
	{
      sieve(); 
      Scanner scan= new Scanner(System.in);
      int testcases=scan.nextInt();
      for(int m=0; m<testcases; m++)
      {
    	  long a =scan.nextLong();
    	  double sqr = Math.sqrt(a);
    	  if((long)sqr == sqr && isPrime[(int)sqr])
    	  {
    		  System.out.println("YES");
    	  }
    	  else System.out.println("NO");
      }
	}
}