#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e3 + 4;
long long int ds(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n = 0, i, k;
  cin >> k;
  for (int i = 19; i < 2 * 10000000 + 1; i++) {
    long long int d = ds(i);
    if (d == 10) {
      n++;
      if (n == k) {
        cout << i;
        return 0;
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int N = (int)1e3 + 4;
long long int ds(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n = 0, i, k;
  cin >> k;
  for (int i = 19; i < 2 * 10000000 + 1; i++) {
    long long int d = ds(i);
    if (d == 10) {
      n++;
      if (n == k) {
        cout << i;
        return 0;
      }
    }
  }
  return 0;
}

```