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: Accepted)

//agc021-a sub
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384

int max(int a,int b){if(a>b){return a;}return b;}
int min(int a,int b){if(a<b){return a;}return b;}
int zt(int a,int b){return max(a,b)-min(a,b);}
int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
int ceil(int a,int b){if(a%b==0){return a/b;}return (a/b)+1;}
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}
int sortfncsj(const void *a,const void *b){if(*(int *)a>*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int sortfnckj(const void *a,const void *b){if(*(int *)a<*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int llsortfncsj(const void *a,const void *b){if(*(long long *)a>*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int llsortfnckj(const void *a,const void *b){if(*(long long *)a<*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int dbsortfncsj(const void *a,const void *b){if(*(double *)a>*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int dbsortfnckj(const void *a,const void *b){if(*(double *)a<*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int strsortfncsj(const void *a,const void *b){return strcmp((char *)a,(char *)b);}
int strsortfnckj(const void *a,const void *b){return strcmp((char *)b,(char *)a);}

int main(void){
    long long x=0,i,j,n,m,k=1,a[262144],b,c,w,r=0,l,t;
    double d;
    char s[262144];
    scanf("%lld",&n);
    //l=strlen(s);
    if(n<10){printf("%lld\n",n);return 0;}
    m=n;
    while(m>=10){
        m/=10;k*=10;x+=9;
    }
    r=x;
    for(i=1;i<10;i++){
        if(i*k+(k-1)<=n){r=x+i;}
    }
    //qsort(a,n,sizeof(int),sortfncsj);
    printf("%lld\n",r);
    return 0;
}