#include <bits/stdc++.h>
using namespace std;
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<T> &v) {
  for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
    out << v[i] << " ";
  out << "\n";
  return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<vector<T> > &v) {
  for (size_t i = 0; i < v.size(); ++i) {
    for (size_t j = 0; j < v[i].size(); ++j) {
      out << v[i][j] << " ";
    }
    out << "\n";
  }
  return out;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> divisors;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      divisors.push_back(i);
      if (n / i != i) {
        divisors.push_back(n / i);
      }
    }
  }
  if (k > divisors.size()) {
    cout << "-1\n";
    return 0;
  }
  sort(divisors.begin(), divisors.end());
  cout << divisors[k - 1] << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<T> &v) {
  for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
    out << v[i] << " ";
  out << "\n";
  return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<vector<T> > &v) {
  for (size_t i = 0; i < v.size(); ++i) {
    for (size_t j = 0; j < v[i].size(); ++j) {
      out << v[i][j] << " ";
    }
    out << "\n";
  }
  return out;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> divisors;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      divisors.push_back(i);
      if (n / i != i) {
        divisors.push_back(n / i);
      }
    }
  }
  if (k > divisors.size()) {
    cout << "-1\n";
    return 0;
  }
  sort(divisors.begin(), divisors.end());
  cout << divisors[k - 1] << "\n";
  return 0;
}

```