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.Collections;
import java.util.Scanner;

public class a {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long n = in.nextLong();
		int k = in.nextInt();
		ArrayList<Long> divisors = new ArrayList<Long>();
		for (long i = 1; i <= Math.sqrt(n); i++) {
			if (n % i == 0) {
				divisors.add(i);
				if (n / i != i)
					divisors.add(n / i);
			}
		}
		Collections.sort(divisors);
		//System.out.println(divisors);
		if (divisors.size() < k) {
			System.out.println(-1);
		} else {
			System.out.println(divisors.get(k - 1));
		}
	}
}