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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define mod 1000000007
typedef long long int ll;
int u(const void *a, const void *b){
    return *(ll*)a-*(ll*)b;
}
int d(const void *a, const void *b){
    return *(ll*)b-*(ll*)a;
}
int z(int a){
  if(a<1)a*=-1;
  return a;
}
int min(int a,int b){
  if(a>b)return b;
  return a;
}
int max(int a,int b){
  if(a>b)return a;
  return b;
}
int gcd(int a,int b){
  if(!b)return a;
  return gcd(b,a%b);
}
int kt(ll a){
  int sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
int ks(ll a){
  int sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
int main(void){
  char s[20];
  scanf("%s",s);
  int flg=0;
  FOR(strlen(s))if(i&&s[i]!='9')flg=1;
  PRN((s[0]-'0')-flg+9*(strlen(s)-1));
  return 0;
}