#include <bits/stdc++.h>
using namespace std;
long long n, m, i, j, k, l, t;
long long a[1000005], b;
bool q = false;
int main() {
  cin >> n >> k;
  for (i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      t++;
      a[t] = i;
    }
    if (i * i == n) q = true;
  }
  if (q == false) {
    if (k <= t)
      cout << a[k] << endl;
    else if (k > t && k <= 2 * t)
      cout << n / a[t - (k - t) + 1] << endl;
    else
      cout << -1 << endl;
  } else {
    if (k <= t)
      cout << a[k] << endl;
    else if (k > t && k <= (2 * t - 1))
      cout << n / a[t - (k - t)] << endl;
    else
      cout << -1 << 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;
long long n, m, i, j, k, l, t;
long long a[1000005], b;
bool q = false;
int main() {
  cin >> n >> k;
  for (i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      t++;
      a[t] = i;
    }
    if (i * i == n) q = true;
  }
  if (q == false) {
    if (k <= t)
      cout << a[k] << endl;
    else if (k > t && k <= 2 * t)
      cout << n / a[t - (k - t) + 1] << endl;
    else
      cout << -1 << endl;
  } else {
    if (k <= t)
      cout << a[k] << endl;
    else if (k > t && k <= (2 * t - 1))
      cout << n / a[t - (k - t)] << endl;
    else
      cout << -1 << endl;
  }
  return 0;
}

```