Problem: You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output

If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples

Input

4 2


Output

2


Input

5 3


Output

-1


Input

12 5


Output

6

Note

In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.PrintWriter;

public class A762 implements Runnable {
	private Scanner in = new Scanner(System.in);
	private PrintWriter out = new PrintWriter(System.out); 

	private long n;
	private int k;

	public static void main(String[] args) {
		new Thread(new A762()).start();
	}

	private void read() {
		n = in.nextLong();
		k = in.nextInt();
		
	}

	private void solve() {

		boolean bool = false;
		
		int s = (int) Math.sqrt(n);
		int count = 0;
		ArrayList<Integer> a = new ArrayList<Integer>();

		for(int i = s; i >= 1; i--) {
			if(n % i == 0) {
				a.add(i);
			}
		}

		if (a.size() >= k) {
			out.println(a.get(a.size() - k));
			bool = true;
		} else {
			count+= a.size();
			for(int i = 0; i < a.size(); i++) {
				if(n%a.get(i) == 0 && n/a.get(i) > s) {
					count++;

					if(count == k){
						out.println(n/a.get(i));
						bool = true;
						break;
					}

				}
			}
		}

		if(!bool){
			out.println("-1");
		}





	}

	private void write() {

	}

	public void run() {
		read();
		solve();
		write();
		out.close();
	} 
}