#include <bits/stdc++.h>
using namespace std;
int sum(long long n) {
  int res = 0;
  while (n != 0) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
int main() {
  long long n, temp;
  cin >> n;
  temp = n;
  string arr = "";
  int k = 0;
  while (temp != 0) {
    char ch = temp % 10 + 48;
    arr += ch;
    temp /= 10;
  }
  int first_dig = 0;
  long long half = 0;
  reverse(arr.begin(), arr.end());
  for (int i = 0; i < arr.size(); i++) {
    if (i == 0) {
      half = half * 10 + arr[i] - 48 - 1;
      first_dig = arr[i] - 48 - 1;
    } else {
      half = half * 10 + 9;
      first_dig += 9;
    }
  }
  long long sec_half = n - half;
  int sec_dig = sum(sec_half);
  cout << first_dig + sec_dig << endl;
}

### Prompt

In cpp, your task is to solve 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 sum(long long n) {
  int res = 0;
  while (n != 0) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
int main() {
  long long n, temp;
  cin >> n;
  temp = n;
  string arr = "";
  int k = 0;
  while (temp != 0) {
    char ch = temp % 10 + 48;
    arr += ch;
    temp /= 10;
  }
  int first_dig = 0;
  long long half = 0;
  reverse(arr.begin(), arr.end());
  for (int i = 0; i < arr.size(); i++) {
    if (i == 0) {
      half = half * 10 + arr[i] - 48 - 1;
      first_dig = arr[i] - 48 - 1;
    } else {
      half = half * 10 + 9;
      first_dig += 9;
    }
  }
  long long sec_half = n - half;
  int sec_dig = sum(sec_half);
  cout << first_dig + sec_dig << endl;
}

```