#include <bits/stdc++.h>
using namespace std;
int primes[3005];
void sieve() {
  memset(primes, 0, sizeof(primes));
  for (int i = 2; i <= 3000; i++) {
    if (primes[i] == 0) {
      for (int j = 2 * i; j <= 3000; j += i) primes[j]++;
    }
  }
  for (int i = 1; i <= 3000; i++)
    if (primes[i] == 0) primes[i]++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  vector<long long> factors;
  ;
  for (long long i = 1; i < sqrt(n); i++) {
    if (n % i == 0) {
      factors.push_back(i);
    }
  }
  sort(factors.begin(), factors.end());
  long long tt = factors.size() * 2;
  if ((long long)sqrt(n) * (long long)sqrt(n) == n) tt++;
  if (k > tt) {
    cout << "-1\n";
    return 0;
  }
  if (factors.size() >= k) {
    cout << factors[k - 1];
  } else {
    if ((long long)sqrt(n) * (long long)sqrt(n) == n && k == factors.size() + 1)
      cout << (long long)sqrt(n);
    else {
      k -= factors.size();
      if ((long long)sqrt(n) * (long long)sqrt(n) == n) k--;
      cout << n / factors[factors.size() - 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;
int primes[3005];
void sieve() {
  memset(primes, 0, sizeof(primes));
  for (int i = 2; i <= 3000; i++) {
    if (primes[i] == 0) {
      for (int j = 2 * i; j <= 3000; j += i) primes[j]++;
    }
  }
  for (int i = 1; i <= 3000; i++)
    if (primes[i] == 0) primes[i]++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  vector<long long> factors;
  ;
  for (long long i = 1; i < sqrt(n); i++) {
    if (n % i == 0) {
      factors.push_back(i);
    }
  }
  sort(factors.begin(), factors.end());
  long long tt = factors.size() * 2;
  if ((long long)sqrt(n) * (long long)sqrt(n) == n) tt++;
  if (k > tt) {
    cout << "-1\n";
    return 0;
  }
  if (factors.size() >= k) {
    cout << factors[k - 1];
  } else {
    if ((long long)sqrt(n) * (long long)sqrt(n) == n && k == factors.size() + 1)
      cout << (long long)sqrt(n);
    else {
      k -= factors.size();
      if ((long long)sqrt(n) * (long long)sqrt(n) == n) k--;
      cout << n / factors[factors.size() - k];
    }
  }
  return 0;
}

```