#include <bits/stdc++.h>
using namespace std;
const int INF = (int)(1u << 31) - 1u;
const int N = 1e5;
const int MOD = 1e9 + 7ll;
const int DX[] = {+1, -1, 0, 0};
const int DY[] = {0, 0, +1, -1};
int digitsSum(int x) {
  int sum = 0;
  while (x) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(NULL);
  int k;
  cin >> k;
  for (int x = 1; k; x++)
    if (!(k -= (digitsSum(x) == 10))) cout << x, exit(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 INF = (int)(1u << 31) - 1u;
const int N = 1e5;
const int MOD = 1e9 + 7ll;
const int DX[] = {+1, -1, 0, 0};
const int DY[] = {0, 0, +1, -1};
int digitsSum(int x) {
  int sum = 0;
  while (x) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(NULL);
  int k;
  cin >> k;
  for (int x = 1; k; x++)
    if (!(k -= (digitsSum(x) == 10))) cout << x, exit(0);
  return 0;
}

```