#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 5;
const long long int Nb = 12345689012345678;
const long double pi = 3.141592653589793;
const long long int md = 1000000007;
void io(bool ok) {
  freopen("input.txt", "r", stdin);
  if (ok) freopen("output.txt", "w", stdout);
}
vector<long long int> factor;
int main() {
  long long int n, k;
  scanf("%lld%lld", &n, &k);
  for (long long int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      factor.push_back(i);
      if (i * i != n) factor.push_back(n / i);
    }
  }
  sort(factor.begin(), factor.end());
  if (k > factor.size())
    printf("-1");
  else
    printf("%lld", factor[k - 1]);
  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 N = 100000 + 5;
const long long int Nb = 12345689012345678;
const long double pi = 3.141592653589793;
const long long int md = 1000000007;
void io(bool ok) {
  freopen("input.txt", "r", stdin);
  if (ok) freopen("output.txt", "w", stdout);
}
vector<long long int> factor;
int main() {
  long long int n, k;
  scanf("%lld%lld", &n, &k);
  for (long long int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      factor.push_back(i);
      if (i * i != n) factor.push_back(n / i);
    }
  }
  sort(factor.begin(), factor.end());
  if (k > factor.size())
    printf("-1");
  else
    printf("%lld", factor[k - 1]);
  return 0;
}

```