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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.StringTokenizer;

/**
 *
 * @author arabtech
 */
public class KThDivisor {

    public static ArrayList<Long> div = new ArrayList<>();

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer line = new StringTokenizer(br.readLine());
        long n = Long.parseLong(line.nextToken());
        long k = Long.parseLong(line.nextToken());
        getDivisors(n);
        if (div.size() < k) {
            System.out.println(-1);
        } else {
            Collections.sort(div);
            System.out.println(div.get((int) k - 1));
        }
    }

    public static void getDivisors(long n) {
        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                if (n / i == i) {
                    div.add((long)i);
                } else {
                    div.add((long)i);
                    div.add((long) n / i);
                }
            }
        }
    }

}