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.*;
public class Divisor
{
public static void main(String...s)
{
	ArrayList <Long> arrLi = new ArrayList<Long>();
	Scanner sc = new Scanner(System.in);
	long n, i;
	n = sc.nextLong();
	int k = sc.nextInt();
	for(i=1; i*i<=n; i++)
	{
		if(n%i==0)
		{
			if(n == i*i)
				arrLi.add(i);
			else
			{
				arrLi.add(i);
				arrLi.add(n/i);
			}
		}
	}
	if(k > arrLi.size())
		System.out.println(-1);
	else
	{
		Collections.sort(arrLi);
		System.out.println(arrLi.get(k-1));
	}
}
}