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 Main {

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

}