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.math.*;
public class IsTPrime {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
boolean prime[] = new boolean[1000001];
for(int i = 2; i < 1000001; i++) prime[i] = true;
for(int i = 2; i <= (int)Math.sqrt(10000000); i++) {
	  if(prime[i]) {
		  for(int j = 2; j * i <= 1000000; j++) {
			  prime[i * j] = false;
		  }
	  }
}
int n = sc.nextInt();
for(int i = 0; i < n; i++) {
	long a = sc.nextLong();
	if(a == 1) {
		System.out.println("NO");
	}
	else {
	int c = (int)Math.sqrt(a);
	long d = (long)c * (long)c;
	if(d != a) {
		System.out.println("NO");
	}
	else {
		int b = (int)Math.sqrt(a);
		if(prime[b]) {
			System.out.println("YES");
		}
		else System.out.println("NO");
	}}
	
}
}
}