#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e7 + 5;
int num[maxn];
long long cnt = 0;
int main() {
  long long n, k;
  cin >> n >> k;
  for (int i = 1; i <= sqrt(n + 0.5); i++) {
    if (n % i == 0) num[cnt++] = i;
  }
  long long s = sqrt(n);
  if (s * s == n) {
    if (cnt * 2 - 1 < k) {
      puts("-1");
      return 0;
    }
    if (cnt >= k)
      cout << num[k - 1] << endl;
    else
      cout << n / num[((cnt - 1) - (k - cnt) + 1) - 1] << endl;
  } else {
    if (cnt * 2 < k) {
      puts("-1");
      return 0;
    }
    if (cnt >= k)
      cout << num[k - 1] << endl;
    else
      cout << n / num[((cnt - 1) - (k - cnt) + 1)] << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int maxn = 1e7 + 5;
int num[maxn];
long long cnt = 0;
int main() {
  long long n, k;
  cin >> n >> k;
  for (int i = 1; i <= sqrt(n + 0.5); i++) {
    if (n % i == 0) num[cnt++] = i;
  }
  long long s = sqrt(n);
  if (s * s == n) {
    if (cnt * 2 - 1 < k) {
      puts("-1");
      return 0;
    }
    if (cnt >= k)
      cout << num[k - 1] << endl;
    else
      cout << n / num[((cnt - 1) - (k - cnt) + 1) - 1] << endl;
  } else {
    if (cnt * 2 < k) {
      puts("-1");
      return 0;
    }
    if (cnt >= k)
      cout << num[k - 1] << endl;
    else
      cout << n / num[((cnt - 1) - (k - cnt) + 1)] << endl;
  }
  return 0;
}

```