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.PrintStream;
import java.util.Scanner;
import java.util.TreeSet;

public class NewKe {
	
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		PrintStream p=System.out;
		try{
			long n=sc.nextLong();
			int k=sc.nextInt();
			TreeSet<Long> ts=new TreeSet<Long>();
			ts.add(new Long(1));
			ts.add(n);
			long nn=(long) Math.sqrt(n);
			for(long i=2;i<=nn;i++){
				if(n%i==0){
					ts.add(i);
					ts.add(n/i);
				}
			}
			if(k>ts.size()){
				p.println(-1);
			}else{
				int time=0;
				for(Long l:ts){
					//p.println("*:"+l);
					time++;
					if(time==k){
						p.println(l);
						break;
					}
				}
			}
		}catch(Exception e){
			e.printStackTrace();
		}finally{
			p.close();
			sc.close();
		}
		
	}
}