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 Solution{  
	
	
	
 public static void main(String[] args) {
     
     Scanner in = new Scanner(System.in);
     
     
     int k,num_d=0;
     long i,n;
    long[] divisor;
    
    
    n=in.nextLong();
    k=in.nextInt();
    
    if(k>44721359){
    	System.out.println(-1);
    	return;
    }
    
    divisor=new long[k+1];
    
    for(i=1L;i*i<=n;i++){
    	if(n%i==0){
    		num_d++;
    		divisor[num_d]=i;
    		if(num_d==k)break;
    	}
    }
    
    if(num_d==k){
    	System.out.println(divisor[k]);
    	return;
    }
    
    else if((i-1)*(i-1)==n){
    	if(2*num_d-1<k){
    		System.out.println(-1);
    		return;
    	}
    	else{
    		int l=num_d;
    		for(i=l-1;i>=2*l-k;i--){
    			num_d++;
    			divisor[num_d]=(n/divisor[(int)i]);
    		}
    	}
    	System.out.println(divisor[k]);
    	return;
    }
    
    else if(i*i!=n){
    	if(2*num_d<k){
    		System.out.println(-1);
    		return;
    	}
    	else{
    		int l=num_d;
    		for(i=l;i>=2*l+1-k;i--){
    			num_d++;
    			divisor[num_d]=(n/divisor[(int)i]);
    		}
    	}
    	System.out.println(divisor[k]);
    	return;
    }
    
    
 }
}