#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
vector<long long int> v;
void f(long long int n) {
  for (long long int i = 1; i * i <= n; i++)
    if (n % i == 0) v.push_back(i), v.push_back(n / i);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long int n;
  int k;
  cin >> n >> k;
  f(n);
  sort(v.begin(), v.end());
  v.resize(distance(v.begin(), unique(v.begin(), v.end())));
  if (k > (int)v.size()) return cout << -1 << endl, 0;
  cout << v[k - 1] << endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
vector<long long int> v;
void f(long long int n) {
  for (long long int i = 1; i * i <= n; i++)
    if (n % i == 0) v.push_back(i), v.push_back(n / i);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long int n;
  int k;
  cin >> n >> k;
  f(n);
  sort(v.begin(), v.end());
  v.resize(distance(v.begin(), unique(v.begin(), v.end())));
  if (k > (int)v.size()) return cout << -1 << endl, 0;
  cout << v[k - 1] << endl;
  return 0;
}

```