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.*;

public class Main {	
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		
		long n = sc.nextLong();
		int k = sc.nextInt();
		List<Long> list1 = new LinkedList<>();
		List<Long> list2 = new LinkedList<>();
		
		for(long i=1;i*i<=n;i++){
			if(n%i==0){
				list1.add(i);
				if(i*i != n)
					list2.add(0, n/i);
			}
		}
		
		list1.addAll(list2);
		
		if(list1.size() >= k)
			System.out.println(list1.get(k-1));
		else
			System.out.println(-1);
	}
}