#include <bits/stdc++.h>
using namespace std;
bool n(long long n) {
  long long s = 0;
  while (n > 0) {
    s += (n % 10);
    n /= 10;
  }
  if (s == 10) return true;
  return false;
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long k = 0;
  long long j = 19;
  vector<long long> c;
  while (k != 10000) {
    if (n(j)) c.push_back(j), k++;
    j++;
  }
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    cout << c[n - 1] << "\n";
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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 n(long long n) {
  long long s = 0;
  while (n > 0) {
    s += (n % 10);
    n /= 10;
  }
  if (s == 10) return true;
  return false;
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long k = 0;
  long long j = 19;
  vector<long long> c;
  while (k != 10000) {
    if (n(j)) c.push_back(j), k++;
    j++;
  }
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    cout << c[n - 1] << "\n";
  }
}

```