#include <bits/stdc++.h>
long long p[16];
void F() {
  p[1] = 10;
  for (int i = 2; i <= 14; i++) p[i] = p[i - 1] * 10;
}
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  F();
  long long n, ans = 0;
  cin >> n;
  if (n < 10) {
    cout << n;
    return 0;
  }
  for (int i = 1; i <= 14; i++) {
    if (n / p[i] == 0) {
      long long a = p[i - 1] - 1;
      long long b = n - a;
      ostringstream str1, str2;
      str1 << a;
      str2 << b;
      string sa = str1.str();
      string sb = str2.str();
      for (auto x : sa) ans += x - '0';
      for (auto x : sb) ans += x - '0';
      cout << ans;
      return 0;
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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>
long long p[16];
void F() {
  p[1] = 10;
  for (int i = 2; i <= 14; i++) p[i] = p[i - 1] * 10;
}
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  F();
  long long n, ans = 0;
  cin >> n;
  if (n < 10) {
    cout << n;
    return 0;
  }
  for (int i = 1; i <= 14; i++) {
    if (n / p[i] == 0) {
      long long a = p[i - 1] - 1;
      long long b = n - a;
      ostringstream str1, str2;
      str1 << a;
      str2 << b;
      string sa = str1.str();
      string sb = str2.str();
      for (auto x : sa) ans += x - '0';
      for (auto x : sb) ans += x - '0';
      cout << ans;
      return 0;
    }
  }
  return 0;
}

```