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.ArrayList;
import java.util.Scanner;

public class CodeForce
{
	public static void main(String args[])
	{
		Scanner input = new Scanner(System.in);
		long n = input.nextLong();
		int k = input.nextInt();
		ArrayList<Long> list = new ArrayList<>();
		list.add(0L);
		boolean square = false;
		long t = (long) Math.sqrt(n);
		if (n == t * t)
		{
			square = true;
		}
		int current = 0;
		for (long i = 1; i <= Math.sqrt(n); i++)
		{
			if (n % i == 0)
			{
				current++;
				if (current <= k)
				{
					list.add(i);
				}
			}
		}
		int totalDiv = 2 * current;
		if (square)
		{
			totalDiv--;
		}
		if (totalDiv < k)
		{
			System.out.println(-1);
		} else
		{
			if (current >= k)
			{
				System.out.println(list.get(k));
			} else
			{
				System.out.println(n / list.get(totalDiv + 1 - k));
			}
		}
	}
}