#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int k, i = 0, sum = 10, sum1 = 0, temp, arr[50000], r, j = 0;
  cin >> k;
  for (i = 0; i < 9000000; i++) {
    sum = sum + 9;
    temp = sum;
    while (temp != 0) {
      r = temp % 10;
      sum1 = sum1 + r;
      temp = temp / 10;
    }
    if (sum1 == 10) {
      arr[j] = sum;
      j++;
    }
    sum1 = 0;
  }
  for (int l = 0; l < 90000; l++) {
    if (l + 1 == k) {
      cout << arr[l] << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
int main() {
  long long int k, i = 0, sum = 10, sum1 = 0, temp, arr[50000], r, j = 0;
  cin >> k;
  for (i = 0; i < 9000000; i++) {
    sum = sum + 9;
    temp = sum;
    while (temp != 0) {
      r = temp % 10;
      sum1 = sum1 + r;
      temp = temp / 10;
    }
    if (sum1 == 10) {
      arr[j] = sum;
      j++;
    }
    sum1 = 0;
  }
  for (int l = 0; l < 90000; l++) {
    if (l + 1 == k) {
      cout << arr[l] << endl;
    }
  }
  return 0;
}

```