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.HashSet;
import java.util.Scanner;
import java.io.*; 
import java.util.*; 

public class T_Primes_230B {
	
	static long[] primes = new long[1000001];
	static boolean[] composite = new boolean[1000001];
	static int counter = 0;
	
	public static void main(String[] args0) throws Exception {
		@SuppressWarnings("resource")
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = Integer.parseInt(br.readLine()); 
		
		HashSet<Long> tprimes = new HashSet<Long>();
		
		sieve(1000000);
		for(int i = 0; i < counter; i++) {
			tprimes.add((long) (primes[i] * primes[i]));
		}
		
		StringTokenizer st = new StringTokenizer(br.readLine()); 
		
		for(int i = 0; i < n; i++) {
			if(tprimes.contains(Long.parseLong(st.nextToken()))) {
				pw.println("YES");
			} else {
				pw.println("NO");
			}
		}
		pw.close();
		
	}
	
	public static void sieve(long n) {
		composite[0] = true;
		composite[1] = true;
		for(int i = 2; i * i <= n; i++) {
			if(!composite[i]) {
				for(int j = i + i; j <= n; j += i) {
					composite[j] = true;
				}
			}
		}

		for(int i = 2; i <= n; i++) {
			if(!composite[i]) primes[counter++] = i;
		}
	}
}