#include <bits/stdc++.h>
using namespace std;
bool sumofnumbers(long long a) {
  int ans = 0;
  while (a) {
    ans = ans + a % 10;
    a = a / 10;
  }
  if (ans == 10) return true;
  return false;
}
int main() {
  int k;
  cin >> k;
  long long* arr = new long long[k];
  arr[0] = 19;
  for (int i = 1; i < k; i++) {
    long long a = 9 + arr[i - 1];
    arr[i] = 9 + arr[i - 1];
    while (!sumofnumbers(a)) {
      arr[i] = arr[i] + 9;
      a = arr[i];
    }
  }
  cout << arr[k - 1];
  return 0;
}

### 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;
bool sumofnumbers(long long a) {
  int ans = 0;
  while (a) {
    ans = ans + a % 10;
    a = a / 10;
  }
  if (ans == 10) return true;
  return false;
}
int main() {
  int k;
  cin >> k;
  long long* arr = new long long[k];
  arr[0] = 19;
  for (int i = 1; i < k; i++) {
    long long a = 9 + arr[i - 1];
    arr[i] = 9 + arr[i - 1];
    while (!sumofnumbers(a)) {
      arr[i] = arr[i] + 9;
      a = arr[i];
    }
  }
  cout << arr[k - 1];
  return 0;
}

```