#include <bits/stdc++.h>
using namespace std;
map<int, long long> mp;
int main(void) {
  long long k, n;
  int cc = 0;
  cin >> n >> k;
  long long gg = sqrt(n);
  for (long long i = 1; i <= gg; i++) {
    if (n % i == 0) {
      mp[cc++] = i;
    }
  }
  int ok = 0, num = 0;
  if (n == gg * gg) {
    ok = 1;
  }
  num = cc * 2 - ok;
  if (num < k) {
    cout << -1 << endl;
  } else {
    if (k <= cc) {
      cout << mp[k - 1] << endl;
    } else {
      int x = num - k;
      long long y = n / mp[x];
      cout << y << endl;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
map<int, long long> mp;
int main(void) {
  long long k, n;
  int cc = 0;
  cin >> n >> k;
  long long gg = sqrt(n);
  for (long long i = 1; i <= gg; i++) {
    if (n % i == 0) {
      mp[cc++] = i;
    }
  }
  int ok = 0, num = 0;
  if (n == gg * gg) {
    ok = 1;
  }
  num = cc * 2 - ok;
  if (num < k) {
    cout << -1 << endl;
  } else {
    if (k <= cc) {
      cout << mp[k - 1] << endl;
    } else {
      int x = num - k;
      long long y = n / mp[x];
      cout << y << endl;
    }
  }
  return 0;
}

```