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.*;
import java.util.StringTokenizer;
public class CF0230B
{
 public static void main(String[] helloWorld) throws IOException
 {
  boolean[] all = new boolean[1000001];
  for (int i = 2; i <= 1000; i ++)
   if (!all[i])
    for (int j = i * i; j <= 1000000; j += i)
     all[j] = true;
  all[1] = true;
  BufferedReader sIn = new BufferedReader(new InputStreamReader(System.in));
  PrintStream sOut = new PrintStream(new BufferedOutputStream(System.out));
  int n = Integer.parseInt(sIn.readLine());
  StringTokenizer sT = new StringTokenizer(sIn.readLine());
  for (int i = 0; i < n; i ++)
  {
   long l = Long.parseLong(sT.nextToken());
   long root = Math.round(Math.sqrt(l));
   sOut.println(!all[(int) root] && root * root == l ? "YES" : "NO");
  }
  sOut.close();
 }
}