#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
int a[maxn], b[maxn];
int n;
inline long long dfs(long long t) {
  long long n = t;
  int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  if (sum == 10)
    return 1;
  else
    return 0;
}
int main() {
  scanf("%d", &n);
  int l = 0;
  for (int i = 1; i < 10800101; i++) {
    if (dfs(i)) b[++l] = i;
  }
  printf("%d\n", b[n]);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int maxn = 100000 + 10;
int a[maxn], b[maxn];
int n;
inline long long dfs(long long t) {
  long long n = t;
  int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  if (sum == 10)
    return 1;
  else
    return 0;
}
int main() {
  scanf("%d", &n);
  int l = 0;
  for (int i = 1; i < 10800101; i++) {
    if (dfs(i)) b[++l] = i;
  }
  printf("%d\n", b[n]);
  return 0;
}

```