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.math.BigInteger;
import java.util.*;
public class Tprimes {
	public static boolean[] num=new boolean[1000001];
	public static boolean check(long d){
		double sq = Math.pow((double)d,0.5);
		if(Math.floor(sq) != sq) 
			return false;
		if(num[(int)sq]) 
			return false;
		return true;
	}
	public static void main(String[] args) {
		Scanner S= new Scanner(System.in);
		int n=S.nextInt();
		num[1]=true;
		for(int i = 2; i <= 1000; ++i){
			  if(num[i] == false){
				  for(int j = 2; i*j <= 1000000; ++j){
					  num[i*j] = true;
				  }
			  }
		  }
		for(int i=0;i<n;i++) {
			long s=S.nextLong();
			if(s==1) {
				System.out.println("NO");
			}else {
				if(check(s)) {
					System.out.println("YES");
				}
				else {
					System.out.println("NO");
				}
			}
		}
	}
}