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 kthDiv{
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);

		double n = input.nextDouble();
		double k = input.nextDouble();

		ArrayList <Long> set = new ArrayList<Long>();

		// System.out.println(Math.sqrt(n));
		for(double i = 1; i*i<=n; i++){
			// System.out.println(i);
			if(n%i == 0){

				if((long)i == (long)(n/i)){
					set.add((long)i);
				}
				else{
					set.add((long)i);
					set.add((long)(n/i));
				}
					
			}
		}

		Collections.sort(set);
		// System.out.println(set);

		long x = 0;
		long ans = -1;
		
		for(Long en: set){
			// System.out.println(en);
			x++;
			if(x == k){
				ans = en;
				break;
			}
		}

		System.out.println(ans);
	}
}