Problem:
Score : 300 points

Problem Statement

Find the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Constraints

1\leq N \leq 10^{16}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Sample Input 1

100

Sample Output 1

18

For example, the sum of the digits in 99 is 18, which turns out to be the maximum value.

Sample Input 2

9995

Sample Output 2

35

For example, the sum of the digits in 9989 is 35, which turns out to be the maximum value.

Sample Input 3

3141592653589793

Sample Output 3

137

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int
main(int argc, char *argv[])
{
  char s[20] = {}; int ans, k;
  scanf("%s", s);
  int i = 0, d = 0;
  while(s[i] == 9) i++;
  while(s[d] != 0) d++;
  if(d == 1) ans = s[0] - '0';
  else{
    if(i > 0){
      ans = 9 * d - 1;
    }
    else{
      k = 1;
      while(s[k] == '9') k++;
      if(k == d){ans = s[0] - '0' + 9 * (d - 1); printf("1\n");}
      else ans = s[0] - '0' - 1 + 9 * (d - 1);
    }
  }

  printf("%d\n", ans);

  return 0;
}