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.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Main{
	public static void main(String[] args){
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String firstLine = "";
        try {
            firstLine = br.readLine();
        } catch (Exception e) {
        }
        String[] strs = firstLine.trim().split("\\s+");
        double n = Double.parseDouble(strs[0]);
        double k = Double.parseDouble(strs[1]);
        ArrayList<Double> ldiv = new ArrayList<Double>();
        ArrayList udiv = new ArrayList<Double>();
        double ks = 0D;
        double div = 1D;
        double limit = Math.sqrt(n);
        while (div <= limit) {
            if (n % div == 0) {
                ks++;
                if (Double.valueOf((n / div)) != div) {
                    ldiv.add(div);
                    udiv.add(0, n / div);
                } else {
                    ldiv.add(div);
                }
                if (ks == k) {
                    break;
                }
            }
            div++;
        }
        ldiv.addAll(udiv);
        if (ldiv.size() < k) {
            System.out.println("-1");
        } else {
            System.out.println( ldiv.get((int) k - 1).longValue());
        }
	}
}