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.*;
import java.math.*;
import java.lang.*;

/**
 * Created by vishakha on 07-11-2016.
 */
public class Problem {
    private static int M = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        int k = sc.nextInt();
        divisors(n, k);
    }

    public static void divisors(long n, int k) {
        ArrayList<Long> v = new ArrayList<Long>();
        for (long i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                if (n / i == i) { // check if divisors are equal
                    v.add(i);
                } else {
                    v.add(i);
                    v.add(n / i);
                    // push the second divisor in the vector
                }
            }
        }

        Collections.sort(v);
        if (v.size() >= k) System.out.println(v.get(k - 1));
        else System.out.println(-1) ;
    }
}