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.PrintWriter;
import java.util.BitSet;
import java.util.Scanner;

public class CF_230B {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		BitSet sieve = new BitSet(1_000_000);
		sieve.set(1);
		
		for(int i=2;i<=1_000;i++) {
			if(sieve.get(i))
				continue;
			for(int j=i*i;j<=1_000_000;j+=i)
				sieve.set(j);
		}
		int n = scan.nextInt();
		PrintWriter out = new PrintWriter(System.out);
		for(int i=0;i<n;i++) {
			long num = scan.nextLong();
			double roughsqrt = Math.sqrt(num);
			
			int cd1 = (int)Math.floor(roughsqrt);
			int cd2 = (int)Math.ceil(roughsqrt);
			int sqrt = -1;
			if(cd1 * (long)cd1 == num)
				sqrt = cd1;
			if(cd2 * (long)cd2 == num)
				sqrt = cd1;
			if(sqrt == -1 || sieve.get(sqrt)) {
				out.println("NO");
				continue;
			}
			out.println("YES");
		}
		out.close();
		
	}
}