CodeForces762A

A. k-th divisor

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

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.

 //2017.02.01
 #include <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 int divisor[];

 int main()
 {
     long long n, k;
     while(cin>>n>>k)
     {
         long long ans = -;
         int cnt = ;
         bool fg = false;
         for(long long i = ; i*i <= n; i++)
         {
             if(n%i==)divisor[++cnt] = i;
             if(i*i == n)fg = true;
         }
         if(fg && k > *cnt-)cout<<-<<endl;
         else if(k > *cnt)cout<<-<<endl;
         else if(k<=cnt)cout<<divisor[k]<<endl;
         else{
             if(!fg)cout<<(n/divisor[*cnt+-k])<<endl;
             else cout<<(n/divisor[*cnt-k])<<endl;
         }
     }

     return ;
 }