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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class kthdivisor {

	public static void main(String[] args) throws Exception {
		BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter w = new PrintWriter(System.out);
				
		//BufferedReader r = new BufferedReader(new FileReader("tallbarn.in"));
        //PrintWriter w = new PrintWriter(new BufferedWriter(new FileWriter("tallbarn.out")));
        

		StringTokenizer st = new StringTokenizer(r.readLine());
		long n = Long.parseLong(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		
		ArrayList<Long> factors = new ArrayList<Long>();
		int counter = 0;
		int answer = -1;
		for (int i = 1; i < (int) Math.sqrt(n) + 1; i++) {
			if (n % (long) i == 0) {
				factors.add((long) i);
				if (n / (long) i != (long) i)
					factors.add(n / (long) i);
			}
		}
		Collections.sort(factors);

		if (factors.size() <= k - 1) System.out.println(-1);
		else System.out.println(factors.get(k - 1));
	}

}