#include <bits/stdc++.h>
using namespace std;
long long int sum(long long int n) {
  long long int s2um = 0;
  while (n > 0) {
    s2um += n % 10;
    n = n / 10;
  }
  return s2um;
}
long long int answer(long long int n) {
  long long int x = 0, y = 0;
  while (10 * y + 9 < n) {
    y = 10 * y + 9;
    x = x + 9;
  }
  long long int m = n - y;
  x += sum(m);
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  long long int n;
  cin >> n;
  cout << answer(n);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

You are given a positive integer n.

Let S(x) be sum of digits in base 10 representation of x, for example, S(123) = 1 + 2 + 3 = 6, S(0) = 0.

Your task is to find two integers a, b, such that 0 ≤ a, b ≤ n, a + b = n and S(a) + S(b) is the largest possible among all such pairs.

Input

The only line of input contains an integer n (1 ≤ n ≤ 10^{12}).

Output

Print largest S(a) + S(b) among all pairs of integers a, b, such that 0 ≤ a, b ≤ n and a + b = n.

Examples

Input

35


Output

17


Input

10000000000


Output

91

Note

In the first example, you can choose, for example, a = 17 and b = 18, so that S(17) + S(18) = 1 + 7 + 1 + 8 = 17. It can be shown that it is impossible to get a larger answer.

In the second test example, you can choose, for example, a = 5000000001 and b = 4999999999, with S(5000000001) + S(4999999999) = 91. It can be shown that it is impossible to get a larger answer.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int sum(long long int n) {
  long long int s2um = 0;
  while (n > 0) {
    s2um += n % 10;
    n = n / 10;
  }
  return s2um;
}
long long int answer(long long int n) {
  long long int x = 0, y = 0;
  while (10 * y + 9 < n) {
    y = 10 * y + 9;
    x = x + 9;
  }
  long long int m = n - y;
  x += sum(m);
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  long long int n;
  cin >> n;
  cout << answer(n);
  return 0;
}

```