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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;


public class Tprime {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
		int n = Integer.parseInt(in.readLine());
		String[] temp = in.readLine().split(" ");
		long[] t = new long[n];
		for(int i = 0;i<n;i++) {
			t[i] = Long.parseLong(temp[i]);
		}
		boolean[] shieves = new boolean[1000001];
		Arrays.fill(shieves, true);
		shieves[0] = false;
		shieves[1] = false;
		for(int i = 2;i*i<shieves.length;i++) {
			if(shieves[i]) {
				for(int j = i;i*j<shieves.length;j++) {
					shieves[i*j] = false;
				}
			}
		}
		for(int i = 0;i<t.length;i++) {
			double sq = Math.sqrt(t[i]);
			if(sq - Math.floor(sq) > 0) {
				out.write("NO\n");
			} else {
				if(shieves[(int)sq]) {
					out.write("YES\n");
				} else {
					out.write("NO\n");
				}
			}
		}
		out.flush();
		
	}
}