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.*;
public class Main
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int[] arr=new int[1000001];
		for(int i=3;i<1000001;i+=2)
		{
		    arr[i]=1;
		}
			for(int i=3;i<=1000;i+=2)
		{
		    if(arr[i]==1)
		    {
		        for(int j=i*i;j<1000001;j+=i)
		        {
		            arr[j]=0;
		        }
		    }
		    
		}
	
		arr[2]=1;
		arr[0]=arr[1]=0;
		int t=sc.nextInt();
		for(int o=0;o<t;o++)
		{
		    long n=sc.nextLong();
		    if(Math.sqrt(n)==(int)Math.sqrt(n)&&arr[(int)Math.sqrt(n)]==1)System.out.println("YES");
		    else{
		        System.out.println("NO");
		    }
		}
		
		
		
		
		
		
	}
}