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 T_Prime {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		long num = input.nextLong();
		long[] arr = new long[(int) num];
		long var;
		for(long i=0;i<num;i++)
		{
			arr[(int) i] = input.nextLong();
		}
		long[] prime = new long[10000001];
		prime[0]=1;
		prime[1]=1;
		for(long i=2;i*i<=10000000;i++)
		{
			if(prime[(int) i]==0)
			{
				for(long j=i;i*j<=1000000;j++)
				{
					prime[(int) (i*j)]=1;
				}
			}
		}
		for(long i=0;i<num;i++)
		{
			var = (long) Math.sqrt(arr[(int) i]);
			
			if(var*var==arr[(int) i])
			{
				if(prime[(int) var]==0)
				{
					System.out.println("YES");
				}
				else
				{
					System.out.println("NO");
				}
			}
			else
			{
				System.out.println("NO");
			}
		}
	}
}