#include <bits/stdc++.h>
using namespace std;
long long n, k, i, j, num = 0, sq;
static int ans[10000000];
void planb() {
  for (i = 1, j = 1; i <= sq; i++) {
    if (n % i == 0) ans[j++] = i;
  }
  if (ans[k] != 0)
    cout << ans[k] << endl;
  else if (k <= j * 2 - 2)
    cout << n / ans[2 * j - k - 1] << endl;
  else
    cout << "-1" << endl;
}
int main() {
  cin >> n >> k;
  if (k > 10000000) {
    cout << "-1" << endl;
    return 0;
  }
  sq = sqrt((double)n);
  if (sq * sq == n) {
    for (i = 1, j = 1; i <= sq; i++) {
      if (n % i == 0) ans[j++] = i;
    }
    if (ans[k] != 0)
      cout << ans[k] << endl;
    else if (k <= j * 2 - 3)
      cout << n / ans[j * 2 - 2 - k] << endl;
    else
      cout << "-1" << endl;
  } else
    planb();
  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 n, k, i, j, num = 0, sq;
static int ans[10000000];
void planb() {
  for (i = 1, j = 1; i <= sq; i++) {
    if (n % i == 0) ans[j++] = i;
  }
  if (ans[k] != 0)
    cout << ans[k] << endl;
  else if (k <= j * 2 - 2)
    cout << n / ans[2 * j - k - 1] << endl;
  else
    cout << "-1" << endl;
}
int main() {
  cin >> n >> k;
  if (k > 10000000) {
    cout << "-1" << endl;
    return 0;
  }
  sq = sqrt((double)n);
  if (sq * sq == n) {
    for (i = 1, j = 1; i <= sq; i++) {
      if (n % i == 0) ans[j++] = i;
    }
    if (ans[k] != 0)
      cout << ans[k] << endl;
    else if (k <= j * 2 - 3)
      cout << n / ans[j * 2 - 2 - k] << endl;
    else
      cout << "-1" << endl;
  } else
    planb();
  return 0;
}

```