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 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384


int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;} 
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}    //最大公約数
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}     //最小公倍数
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;} //コンビネーション
int nHr(int a,int b){return nCr(a+b-1,b);}       // 重複組み合わせ
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}  //階乗
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;} // a~bまでの階乗
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}   //数字の各位の和
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}  // b進数の各位の和？
int sankaku(int x){return ((1+x)*x)/2;} //三角数 xまでの和

//以下long long ver
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 llnHr(long long a,long long b){return llnCr(a+b-1,b);}
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;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}

//double
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);}


typedef pair<int, int> ii;
typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
#define forr(i,a,b) ; for(int i=(a); i<(b); i++)
#define clean(arr,val) memset(arr,val,sizeof(arr))
#define forn(i,n) forr(i,0,n)
#define PB push_back
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<pll> vpll;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
bool is_prime (int x){
  if (x<=1) return false ;
  for (int i=2 ; i*i<=x ; i++){
    if (x%i==0) return false ;
  }
   return true;
}
string alpha = "abcdefghijklmnopqrstuvwxyz" ;
string ALPHA = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" ;
    
bool chmin (int a,int b){
    if ( a > b ){
    a = b;
    return true ;
    }
    return false ;
}
 bool chmax (int a,int b){
    if ( a < b ){
    a = b;
    return true ;
    }
    return false ;
}   
    
long long dp(long long  n){
  if ( n == 0){
    return 0;
  }
  
  long long s = max ( lldsum ( n) , dp(n-1) ) ;
    return s ;
}
/*CODE START HERE*/
int main() {
  long long n;
  cin >> n ;
  cout << dp(n) << endl; 
}