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.*; 
import java.util.*;
import java.util.Scanner;
public class Main{
  public static void main (String[]args){
    ArrayList <Long> arrList = new ArrayList<Long>();
    Scanner sc = new Scanner (System.in);
    long num = sc.nextLong();
    long k = sc.nextLong();
    for(long i = 1;i*i <= num; i++){
      if(num %i == 0){
        long x = num/i;
        if(i != (x)){
          arrList.add(i);
          arrList.add(x);
        }
        else{
          arrList.add(i);
        }
      }
    }
    Collections.sort(arrList);
    if((int)k > arrList.size()){
      System.out.println("-1");
    }
    else{
      long k1 = k-1;
      System.out.println(arrList.get((int)k1));
    }
  }
}