#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, c, d;
  cin >> a;
  int col = 0;
  long long a1 = a;
  vector<int> b(0);
  while (a > 0) {
    b.push_back(a % 10);
    a = a / 10;
    col++;
  }
  c = b[col - 1] * pow(10, col - 1) - 1;
  d = a1 - c;
  int col1 = 0;
  vector<int> e(0);
  while (c > 0) {
    e.push_back(c % 10);
    c = c / 10;
    col1++;
  }
  vector<int> f(0);
  int col2 = 0;
  while (d > 0) {
    f.push_back(d % 10);
    d = d / 10;
    col2++;
  }
  int s1 = 0;
  int s2 = 0;
  while (col1 > 0) {
    s1 = s1 + e[col1 - 1];
    col1--;
  }
  while (col2 > 0) {
    s2 = s2 + f[col2 - 1];
    col2--;
  }
  cout << s1 + s2;
  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;
int main() {
  long long a, c, d;
  cin >> a;
  int col = 0;
  long long a1 = a;
  vector<int> b(0);
  while (a > 0) {
    b.push_back(a % 10);
    a = a / 10;
    col++;
  }
  c = b[col - 1] * pow(10, col - 1) - 1;
  d = a1 - c;
  int col1 = 0;
  vector<int> e(0);
  while (c > 0) {
    e.push_back(c % 10);
    c = c / 10;
    col1++;
  }
  vector<int> f(0);
  int col2 = 0;
  while (d > 0) {
    f.push_back(d % 10);
    d = d / 10;
    col2++;
  }
  int s1 = 0;
  int s2 = 0;
  while (col1 > 0) {
    s1 = s1 + e[col1 - 1];
    col1--;
  }
  while (col2 > 0) {
    s2 = s2 + f[col2 - 1];
    col2--;
  }
  cout << s1 + s2;
  return 0;
}

```