#include <bits/stdc++.h>
using namespace std;
struct node {
  int data;
  node *next;
};
node *create_node() { return 0; }
int main() {
  string s;
  long long int num = 0, i, t, sum = 0, num1 = 0, l, rem;
  cin >> s;
  for (i = 0; i < s.length(); i++) {
    t = s[i] - '0';
    num = num * 10 + t;
    if (i < 1) {
      sum += t - 1;
      num1 = num1 * 10 + (t - 1);
    } else {
      sum += 9;
      num1 = num1 * 10 + 9;
    }
  }
  l = num - num1;
  while (l != 0) {
    rem = l % 10;
    sum += rem;
    l = l / 10;
  }
  cout << sum;
  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;
struct node {
  int data;
  node *next;
};
node *create_node() { return 0; }
int main() {
  string s;
  long long int num = 0, i, t, sum = 0, num1 = 0, l, rem;
  cin >> s;
  for (i = 0; i < s.length(); i++) {
    t = s[i] - '0';
    num = num * 10 + t;
    if (i < 1) {
      sum += t - 1;
      num1 = num1 * 10 + (t - 1);
    } else {
      sum += 9;
      num1 = num1 * 10 + 9;
    }
  }
  l = num - num1;
  while (l != 0) {
    rem = l % 10;
    sum += rem;
    l = l / 10;
  }
  cout << sum;
  return 0;
}

```