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.*;
public class B230 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		long m;
		boolean primes[] = new boolean[1000000];
		
		Arrays.fill(primes, true);
		primes[0] = false;
		primes[1] = false;
		for(int i = 2; i < 1000000; i ++)
		{
			if(primes[i] == false)
				continue;
			for(long j = (long)i * i; (long) j < 1000000; j += i)
				primes[(int)j] = false;
		}
		
		for(int i=0; i<n; i++)
		{
			m = in.nextLong();
			if(m%2 != 0)
			{
				double sq = Math.sqrt(m);
				if(sq != (long) sq)
					System.out.println("NO");
				else
					System.out.println(primes[(int)sq] ? "YES" : "NO");
			}
			else if(m == 4)
				System.out.println("YES");
			else 
				System.out.println("NO");
		}
	}

}