#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  long long temp = n;
  long long c = 0;
  while (temp != 0) {
    temp /= 10;
    c++;
  }
  long long ans;
  long long num1;
  num1 = ((n / (long long)pow(10, c - 1)) - 1);
  ans = num1 + (c - 1) * 9;
  long long nm1 = 0;
  for (long long i = 0; i <= c - 2; i++) nm1 += 9 * (long long)(pow(10, i));
  nm1 += num1 * (long long)(pow(10, c - 1));
  long long nm2 = n - nm1;
  while (nm2 != 0) {
    ans += nm2 % 10;
    nm2 /= 10;
  }
  cout << ans;
  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 main() {
  long long n;
  cin >> n;
  long long temp = n;
  long long c = 0;
  while (temp != 0) {
    temp /= 10;
    c++;
  }
  long long ans;
  long long num1;
  num1 = ((n / (long long)pow(10, c - 1)) - 1);
  ans = num1 + (c - 1) * 9;
  long long nm1 = 0;
  for (long long i = 0; i <= c - 2; i++) nm1 += 9 * (long long)(pow(10, i));
  nm1 += num1 * (long long)(pow(10, c - 1));
  long long nm2 = n - nm1;
  while (nm2 != 0) {
    ans += nm2 % 10;
    nm2 /= 10;
  }
  cout << ans;
  return 0;
}

```