#include <bits/stdc++.h>
using namespace std;
int const MOD = 1e9 + 7;
map<long long int, long long int> mp;
void solve(long long int a) {
  long long int b, c = 1, d = 0, sum = 0, p, x;
  while (1) {
    b = 9 * c;
    d = d + b;
    if (d > a) {
      d = d - b;
      break;
    }
    c = c * 10;
  }
  p = a - d;
  while (1) {
    if (p == 0) {
      break;
    }
    x = p % 10;
    sum = sum + x;
    p = p / 10;
  }
  while (1) {
    if (d == 0) {
      break;
    }
    x = d % 10;
    sum = sum + x;
    d = d / 10;
  }
  cout << sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, i, j, f, temp, x = 0, u;
  cin >> a;
  solve(a);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 const MOD = 1e9 + 7;
map<long long int, long long int> mp;
void solve(long long int a) {
  long long int b, c = 1, d = 0, sum = 0, p, x;
  while (1) {
    b = 9 * c;
    d = d + b;
    if (d > a) {
      d = d - b;
      break;
    }
    c = c * 10;
  }
  p = a - d;
  while (1) {
    if (p == 0) {
      break;
    }
    x = p % 10;
    sum = sum + x;
    p = p / 10;
  }
  while (1) {
    if (d == 0) {
      break;
    }
    x = d % 10;
    sum = sum + x;
    d = d / 10;
  }
  cout << sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, i, j, f, temp, x = 0, u;
  cin >> a;
  solve(a);
  return 0;
}

```