#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e7 * 3 + 10;
long long n, k, q[MAXN], ans;
int main() {
  cin >> n >> k;
  long long t = sqrt(n);
  for (long long i = 1; i <= t; i++)
    if (n % i == 0) q[++ans] = i;
  for (long long i = t - (t * t == n); i >= 1; i--)
    if (n % i == 0) q[++ans] = n / i;
  if (ans < k)
    cout << -1;
  else
    cout << q[k];
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int MAXN = 1e7 * 3 + 10;
long long n, k, q[MAXN], ans;
int main() {
  cin >> n >> k;
  long long t = sqrt(n);
  for (long long i = 1; i <= t; i++)
    if (n % i == 0) q[++ans] = i;
  for (long long i = t - (t * t == n); i >= 1; i--)
    if (n % i == 0) q[++ans] = n / i;
  if (ans < k)
    cout << -1;
  else
    cout << q[k];
  return 0;
}

```