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.SortedSet;  
import java.util.TreeSet;  
import java.util.*;
public class Main{
  public static void main(String[]args){
    Scanner i=new Scanner(System.in);
    long n=i.nextLong();
    int k = i.nextInt();
    
    ArrayList<Long> nums = new ArrayList<Long>();

    for(long x=1;x*x<=n;x++){
      if(n%x==0){
        nums.add(x);
        if(n/x!=x)
            nums.add(n/x);
      }
    }
    
    if(k>nums.size()){
        System.out.println(-1);
        return ;
    }
    Collections.sort(nums);
    
    System.out.println(nums.get(k-1));
    
    
  }
  
}