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 _762A {
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		long n = sc.nextLong();
		int k = sc.nextInt();
		List<Long> answer = new ArrayList<Long>();
		for(long i=1;i*i<=n;i++)
		{
			if (n%i==0)
			{
				answer.add(i);
				if (i*i!=n)
				{
					answer.add(n/i);
				}
			}
			
			
		}
		Collections.sort(answer);
		if (k>answer.size()){
			System.out.println(-1);
		}
		else
		{
			System.out.println(answer.get(k-1));
		}
		/*
		int i = 1;
		int[] arr = new int[k];
		int cnt = 0;
		while (i < (int)Math.sqrt(n)+1)
		{
			if (n%i==0){
				arr[cnt++]=i;
			}
			if(cnt>=k)
			{
				break;
			}
			i+=1;
		}
		if (k>cnt*2)
			System.out.println(-1);
		else{
			if (k>cnt){
				//System.out.println("Count is "+cnt);
				long val = (n / (arr[(cnt)-(k/2)]));
				System.out.println(val);
			}
			else
			{
				System.out.println(arr[k-1]);
			}
		}
		*/
			
	}
}