#include <bits/stdc++.h>
using namespace std;
int main() {
  unsigned long long int a = 0, f, s2 = 0, t, b, n;
  cin >> n;
  t = n;
  unsigned long long int d = 0;
  while (n != 0) {
    f = n % 10;
    n = n / 10;
    if (n == 0) {
      a = a + pow(10, d) * (f - 1);
    } else {
      a = a + pow(10, d) * 9;
      d++;
    }
  }
  unsigned long long int s1 = d * 9 + f - 1;
  b = t - a;
  while (b > 0) {
    s2 = s2 + b % 10;
    b = b / 10;
  }
  unsigned long long int s = s1 + s2;
  cout << s;
}

### 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>
using namespace std;
int main() {
  unsigned long long int a = 0, f, s2 = 0, t, b, n;
  cin >> n;
  t = n;
  unsigned long long int d = 0;
  while (n != 0) {
    f = n % 10;
    n = n / 10;
    if (n == 0) {
      a = a + pow(10, d) * (f - 1);
    } else {
      a = a + pow(10, d) * 9;
      d++;
    }
  }
  unsigned long long int s1 = d * 9 + f - 1;
  b = t - a;
  while (b > 0) {
    s2 = s2 + b % 10;
    b = b / 10;
  }
  unsigned long long int s = s1 + s2;
  cout << s;
}

```