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.io.*;
import java.util.*;

public class Main {

	static int factors[] = new int[2000];

	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		
		long n = sc.nextLong();
		long k = sc.nextLong();
		
		Vector<Long> divList = new Vector<>();
		long sqrtN = (long) Math.sqrt(n);
		for(long i = 1; i <= sqrtN; i++){
			if(n % i == 0){
				divList.add(i);
				if(i * i != n)
					divList.add(n / i);
			}
		}
		
		Collections.sort(divList);
		
		int size = divList.size();
//		for(Long i : divList){
//			System.out.println(i);
//		}
		
		if(k > size)
			System.out.println("-1");
		else{
			System.out.println(divList.get((int) (k - 1)));
		}
		
		sc.close();
	}

}