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 T_primes {

//	numbers like 4,9,25,49.....have exactly 3 divisors.
	public static void main(String[] args) throws java.lang.Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(br.readLine()); long a[]=new long[n];
		StringTokenizer st=new StringTokenizer(br.readLine());
		for(int i=0; i<n; i++) {
			a[i]=Long.parseLong(st.nextToken());
			if(isPrime((long)Math.sqrt(a[i])) && a[i]==(long)Math.sqrt(a[i])*(long)Math.sqrt(a[i]))
				System.out.println("YES");
			else System.out.println("NO");
		}
	}

	static boolean isPrime(long n) {
		if(n<=1) return false;
		if(n<=3) return true;
		if(n%2==0 || n%3==0) return false;
		for(int i=5; i*i<=n; i+=6) {
			if(n%i==0 || n%(i+2)==0) return false;
		}
		return true;
	}
}