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 A {
	
	public static void main(String[] args)throws IOException 
	{	
		long start=System.nanoTime();
		BufferedReader ob = new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb = new StringBuffer();
		
		String s[]=ob.readLine().split(" ");
		
		long n=Long.parseLong(s[0]);
		int k=Integer.parseInt(s[1]);
		
		long list[]=factors(n);
//		System.out.println(list);
		long end=System.nanoTime();
//		System.out.println("time : "+(end-start)/(1e10));
		if(list.length <k)
		{	
			System.out.println("-1");
			return;
		}
		System.out.println(list[k-1]);
		
	}
	static long[] factors(long n) {
		TreeSet<Long> fact = new TreeSet<>();
		for (int i = 1; i <= Math.sqrt(n); i++) {
			if (n % i == 0) {
				if (n / i == i) {
					fact.add((long)i);
				} else {
					fact.add((long)i);
					fact.add(n / i);
				}
			}
		}
		long a[]=new long[fact.size()];
		int i=0;
		for(long x:fact)
		{
			a[i]=x;
			i++;
		}
		return a;
	}
}