#include <bits/stdc++.h>
using namespace std;
long long s(long long n) {
  long long cnt = 0;
  while (n > 0) {
    cnt += n % 10;
    n /= 10;
  }
  return cnt;
}
long long ex[18];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  long long n1 = n / 2, n2 = n - n1;
  long long sol = 0;
  while (n > 0) {
    long long aux = n % 10;
    long long xp = 0, id = 0;
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        xp = i + j;
        if (n / 10 == 0 && xp > 9) continue;
        if (aux == (xp % 10)) {
          id = max(id, xp);
        }
      }
    }
    n /= 10;
    sol += id;
    if (id > 9) n--;
  }
  cout << sol << '\n';
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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 s(long long n) {
  long long cnt = 0;
  while (n > 0) {
    cnt += n % 10;
    n /= 10;
  }
  return cnt;
}
long long ex[18];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  long long n1 = n / 2, n2 = n - n1;
  long long sol = 0;
  while (n > 0) {
    long long aux = n % 10;
    long long xp = 0, id = 0;
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        xp = i + j;
        if (n / 10 == 0 && xp > 9) continue;
        if (aux == (xp % 10)) {
          id = max(id, xp);
        }
      }
    }
    n /= 10;
    sol += id;
    if (id > 9) n--;
  }
  cout << sol << '\n';
  return 0;
}

```