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.ArrayList;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long n = sc.nextLong();
		int index = sc.nextInt();
		ArrayList<Long> listA = new ArrayList<Long>();
		ArrayList<Long> listB = new ArrayList<Long>();

		for(long i = 1;i*i<=n;i++){
			if(n%i==0){
				if(i*i==n)
					listA.add(i);
				else{
					listA.add(i);
					listB.add(n/i);
				}
			}
		}
		
		if(index>(listA.size()+listB.size()))
			System.out.println(-1);
		else{
			if(index>listA.size())
				System.out.println(listB.get(listB.size()-(index-listA.size()-1)-1));
			else
				System.out.println(listA.get(index-1));
		}
	}
}