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.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class ProblemA {
	static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) throws IOException {
		// kont sheyfha el n 10^5 wel mos7af xD
		Long n = sc.nextLong(), k = sc.nextLong();
		ArrayList<Long> div1 = new ArrayList<>(), div2 = new ArrayList<>();
		for (long i = 1; i * i <= n; i++)
			if (n % i == 0) {
				div1.add(i);
				if (i != n / i)
					div2.add(n / i);
			}
		Collections.sort(div1);
		Collections.sort(div2);
		if (div1.size() + div2.size() < k)
			System.out.println(-1);
		else {
			for (long i = 0; i < div1.size(); i++)
				if (--k == 0) {
					System.out.println(div1.get((int) i));
					return;
				}

			for (long i = 0; i < div2.size(); i++)
				if (--k == 0) {
					System.out.println(div2.get((int) i));

				}
		}
	}
}