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:
//package codeforces;

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

/**
 * Created by heping on 17/1/26.
 */
public class A {


    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        long n = scan.nextLong();
        long k = scan.nextLong();

        List<Long> list = new ArrayList<>();
        for (long i = 1; i * i <= n; i ++) {
            if (n % i == 0) {
                k --;
                list.add(i);
                if (k == 0) {
                    System.out.println(i);
                    return;
                }
            }
        }

        for (int i = list.size() - 1; i >= 0; i --) {
            long v = list.get(i);
            if (v * v == n) continue;
            k --;
            if (k == 0) {
                System.out.println(n / v);
                return;
            }
        }
        System.out.println(-1);
    }
}