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[35000000];

int main()
{
    long long n, k;
    while(cin>>n>>k)
    {
        long long ans = -1;
        int cnt = 0;
        bool fg = false;
        for(long long i = 1; i*i <= n; i++)
        {
            if(n%i==0)divisor[++cnt] = i;
            if(i*i == n)fg = true;
        }
        if(fg && k > 2*cnt-1)cout<<-1<<endl;
        else if(k > 2*cnt)cout<<-1<<endl;
        else if(k<=cnt)cout<<divisor[k]<<endl;
        else{
            if(!fg)cout<<(n/divisor[2*cnt+1-k])<<endl;
            else cout<<(n/divisor[2*cnt-k])<<endl;
        }
    }

    return 0;
}