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 CodeForces762A{
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		long n = input.nextLong();
		int k = input.nextInt();
		if(Math.sqrt(n)*2 < k || (n == 1 && k > 1)){
			System.out.println(-1);
			return;
		}
		ArrayList<Long> arr = new ArrayList<>();
		for(int i = 1;i<=Math.sqrt(n);i++){
			if(n%i == 0){
				arr.add((long)i);
				if(n/i != i){
					arr.add((long)n/i);
				}
			}
		}

		if(arr.size() < k){
			System.out.println(-1);
		}
		else{
			Collections.sort(arr);
			System.out.println(arr.get(k-1));
		}
	}
}