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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;


public class B230 {
	static int MAX = 1000001;
	static boolean[] isPrime = new boolean[MAX];
	
	public static void main(String[] args) throws NumberFormatException, IOException {
		Arrays.fill(isPrime, true);
		isPrime[0] = isPrime[1] = false;
		for(int i = 2; i <= MAX; i++) {
			for(int j = i*2; j < MAX; j+=i) {
				if(isPrime[j]) {
					isPrime[j] = false;
				}
			}
		}
		
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int N = Integer.parseInt(f.readLine());
		String[] split = f.readLine().split(" ");
		for(int i = 0; i < N; i++) {
			if(isT(Long.parseLong(split[i]))) System.out.println("YES");
			else System.out.println("NO");
		}
	}
	
	static boolean isT(long x) {
		if(x==1) return false;
		double sqrt = Math.sqrt(x);
		if(sqrt%1!=0) return false;
//		System.out.println(sqrt);
		if(isPrime[(int)sqrt]) return true;
		return false;
	}
	
}