#include <bits/stdc++.h>
using namespace std;
const int N = 20;
char s[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  bool flag = false;
  int ans = 0;
  for (int i = (n); i >= (1); --i) {
    if (i == 1) {
      ans += s[1] - '0' - flag;
      break;
    }
    if (flag) {
      ans += s[i] - '0' + 9;
      continue;
    }
    if (s[i] == '9') {
      ans += 9;
      continue;
    }
    ans += s[i] - '0' + 10;
    flag = true;
  }
  printf("%d\n", ans);
}

### Prompt

Please formulate 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;
const int N = 20;
char s[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  bool flag = false;
  int ans = 0;
  for (int i = (n); i >= (1); --i) {
    if (i == 1) {
      ans += s[1] - '0' - flag;
      break;
    }
    if (flag) {
      ans += s[i] - '0' + 9;
      continue;
    }
    if (s[i] == '9') {
      ans += 9;
      continue;
    }
    ans += s[i] - '0' + 10;
    flag = true;
  }
  printf("%d\n", ans);
}

```