#include <bits/stdc++.h>
using namespace std;
long long sum_digit(long long a) {
  if (a == 0) return 0;
  return a % 10 + sum_digit(a / 10);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  if (n < 10)
    cout << n;
  else {
    long long b = 9;
    while (b <= n) b = 10 * b + 9;
    b /= 10;
    cout << sum_digit(n - b) + sum_digit(b);
  }
  return 0;
}

### Prompt

Generate a cpp solution to 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;
long long sum_digit(long long a) {
  if (a == 0) return 0;
  return a % 10 + sum_digit(a / 10);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  if (n < 10)
    cout << n;
  else {
    long long b = 9;
    while (b <= n) b = 10 * b + 9;
    b /= 10;
    cout << sum_digit(n - b) + sum_digit(b);
  }
  return 0;
}

```