vector、set 练习 k-th divisor

k-th divisor

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.

Example

Input

4 2

Output

2

Input

5 3

Output

-1

Input

12 5

Output

6
代码实现：

#include<cstdio>
#include <set>
#include<vector>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
    set<long long int>a;
    long long int n,k;
    long long int p1,pp;
    scanf("%lld%lld",&n,&k);
    pp=sqrt(n);
    for(int i=1; i<=pp; i++)
    {
        p1=n%i;
        if(p1==0)
        {
            a.insert(n/i);
            a.insert(i);
        }
    }
    vector<long long int>v;
    insert_iterator<vector<long long int> > in_it(v, v.begin());
    copy(a.begin(), a.end(), in_it);
    //printf("%d
",v.size());
    if(k>v.size())
        printf("-1
");

    else
    {
        printf("%lld
", v[k-1]);
    }

    return 0;
}