#include <bits/stdc++.h>
using namespace std;
std::vector<long long int> vec;
bool breaks(long long int num) {
  long long int ans = 0;
  while (num > 0) {
    ans = ans + (num % 10);
    num /= 10;
  }
  if (ans == 10) return true;
  return false;
}
void call() {
  long long int i = 19;
  while (true) {
    if (vec.size() > 10000) break;
    if (breaks(i)) {
      vec.push_back(i);
    }
    i++;
  }
}
int32_t main() {
  call();
  long long int k;
  std::cin >> k;
  std::cout << vec[k - 1] << "\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

We consider a positive integer perfect, if and only if the sum of its digits is exactly 10. Given a positive integer k, your task is to find the k-th smallest perfect positive integer.

Input

A single line with a positive integer k (1 ≤ k ≤ 10 000).

Output

A single number, denoting the k-th smallest perfect integer.

Examples

Input

1


Output

19


Input

2


Output

28

Note

The first perfect integer is 19 and the second one is 28.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
std::vector<long long int> vec;
bool breaks(long long int num) {
  long long int ans = 0;
  while (num > 0) {
    ans = ans + (num % 10);
    num /= 10;
  }
  if (ans == 10) return true;
  return false;
}
void call() {
  long long int i = 19;
  while (true) {
    if (vec.size() > 10000) break;
    if (breaks(i)) {
      vec.push_back(i);
    }
    i++;
  }
}
int32_t main() {
  call();
  long long int k;
  std::cin >> k;
  std::cout << vec[k - 1] << "\n";
}

```