#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
bool isPowerOfTwo(long long int x) { return x && (!(x & (x - 1))); }
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> vec;
  for (long long int i = 1; i * i <= n; ++i) {
    if (n % i == 0) vec.push_back(i);
  }
  long long int a = sqrt(n);
  if ((a * a) == n) {
    if ((2 * vec.size() - 1) >= k) {
      if (k <= vec.size())
        cout << vec[k - 1];
      else
        cout << n / vec[2 * vec.size() - k - 1];
    } else
      cout << "-1";
  } else {
    if ((2 * vec.size()) >= k) {
      if (k <= vec.size())
        cout << vec[k - 1];
      else
        cout << n / vec[2 * vec.size() - k];
    } else
      cout << "-1";
  }
}
int32_t main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) solve();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
#pragma GCC optimize("Ofast")
using namespace std;
bool isPowerOfTwo(long long int x) { return x && (!(x & (x - 1))); }
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> vec;
  for (long long int i = 1; i * i <= n; ++i) {
    if (n % i == 0) vec.push_back(i);
  }
  long long int a = sqrt(n);
  if ((a * a) == n) {
    if ((2 * vec.size() - 1) >= k) {
      if (k <= vec.size())
        cout << vec[k - 1];
      else
        cout << n / vec[2 * vec.size() - k - 1];
    } else
      cout << "-1";
  } else {
    if ((2 * vec.size()) >= k) {
      if (k <= vec.size())
        cout << vec[k - 1];
      else
        cout << n / vec[2 * vec.size() - k];
    } else
      cout << "-1";
  }
}
int32_t main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) solve();
  return 0;
}

```