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.text.*;
import java.math.*;
import java.util.Map;
import java.io.PrintWriter;
import java.util.Map.Entry;
import java.util.regex.*;

public class Solution {
    public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      long n=sc.nextLong();
      long l=sc.nextLong();
      ArrayList<Long> al=new ArrayList<Long>();
      for(long i=1;i*i<=n;i++){
        if (n%i==0)
			{
				al.add(i);
				if (i*i!=n)
				{
					al.add(n/i);
				}
			}
      }
      Collections.sort(al);
      if(al.size()<l)
        System.out.println("-1");
      else
        System.out.println(al.get((int)l-1));
  }
public static int gcd(int p, int q) {
          if (q == 0)
            return p;
          return gcd(q, p % q);
        }

public static boolean isprime(int number){
            for(int i=2; i<=number/2; i++){
                if(number % i == 0){
                    return false;
                }
            }
            return true;
          }
}