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