#include <bits/stdc++.h>
using namespace std;
int judge(int n) {
  int v[50];
  int p = n;
  int k = 0;
  int i = 0;
  while (p != 0) {
    v[i++] = p % 10;
    p /= 10;
  }
  for (int q = 0; q < i; q++) {
    k += v[q];
  }
  if (k == 10)
    return 1;
  else
    return 0;
}
int k[10009];
int main() {
  int n;
  int o = 1;
  ;
  scanf("%d", &n);
  for (int i = 1; i < 100000000; i++) {
    if (judge(i)) {
      k[o++] = i;
    }
    if (o == n + 1) break;
  }
  printf("%d", k[n]);
}

### 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;
int judge(int n) {
  int v[50];
  int p = n;
  int k = 0;
  int i = 0;
  while (p != 0) {
    v[i++] = p % 10;
    p /= 10;
  }
  for (int q = 0; q < i; q++) {
    k += v[q];
  }
  if (k == 10)
    return 1;
  else
    return 0;
}
int k[10009];
int main() {
  int n;
  int o = 1;
  ;
  scanf("%d", &n);
  for (int i = 1; i < 100000000; i++) {
    if (judge(i)) {
      k[o++] = i;
    }
    if (o == n + 1) break;
  }
  printf("%d", k[n]);
}

```