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.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

/**
 * Created by wvcn3016 on 8/14/2017.
 */
public class SmallestDivisor {

    public static long smDivisor() {
        Scanner s = new Scanner(System.in);
        String[] inputList = s.nextLine().split(" ");
        long lngNumber = Long.parseLong(inputList[0]);
        long lngPosition = Long.parseLong(inputList[1]);
        long lngSquare = (long) Math.sqrt(lngNumber);

        List<Long> lstResults = new ArrayList<Long>();

        for (long i = 1; i <= lngSquare; i++) {
            if (lngNumber % i == 0) {
                lstResults.add(i);
                if (i != lngNumber / i) {
                    lstResults.add(lngNumber / i);
                }
            }
        }
        Collections.sort(lstResults);
        if (lstResults.size() < lngPosition) {
            return -1l;
        }
        return lstResults.get((int) lngPosition - 1);
    }

    public static void main(String[] args) {
        System.out.println(smDivisor());
    }
}