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.io.*;
import java.util.*;
public class Main {
	public static void main (String[] args) {
	    Scanner sc=new Scanner(System.in);
	    int n=sc.nextInt();
	    boolean isP[]=new boolean[(int)1e6+2];
	    isP[1]=true;
	    for(int i=2;i*i<=(int)1e6;i++){
	        if(!isP[i])
	        for(int j=i*i;j<=(int)1e6;j+=i){
	            isP[j]=true;
	        }
	    }
	    for(int i=0;i<n;i++){
	        long x=sc.nextLong();
	        long rt=(long)Math.sqrt(x);
	        if(rt*rt!=x)System.out.println("NO");
	        else if(!isP[(int)rt])System.out.println("YES");
	        else System.out.println("NO");
	    }
	}
}