#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, great = 0;
  cin >> n;
  if (n <= 9) {
    cout << n << endl;
    return 0;
  }
  string s = "9";
  while (true) {
    long long aux = stoll(s);
    if (aux <= n) {
      great = aux;
    } else
      break;
    s += '9';
  }
  int sum = 0;
  string s1 = to_string(great), s2 = to_string(n - great);
  for (auto c : s1) sum += (int)(c - '0');
  for (auto c : s2) sum += (int)(c - '0');
  cout << sum << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int main() {
  long long n, great = 0;
  cin >> n;
  if (n <= 9) {
    cout << n << endl;
    return 0;
  }
  string s = "9";
  while (true) {
    long long aux = stoll(s);
    if (aux <= n) {
      great = aux;
    } else
      break;
    s += '9';
  }
  int sum = 0;
  string s1 = to_string(great), s2 = to_string(n - great);
  for (auto c : s1) sum += (int)(c - '0');
  for (auto c : s2) sum += (int)(c - '0');
  cout << sum << endl;
  return 0;
}

```