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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
  using namespace std;
#define ll long long
#define fw(p) for(int w=0;w<(p);w++)
#define fx(p) for(int x=0;x<(p);x++)
#define fy(p) for(int y=0;y<(p);y++)
#define fz(p) for(int z=0;z<(p);z++)
#define fyg(p,g) for(int y=(g);y<(p);y++)
#define fzg(p,g) for(int z=(g);z<(p);z++)
#define ce(d) cout<<d<<endl;
#define vecp(p) int aa;cin>>aa;(p).push_back(aa);
#define vecpl(p) long long aa;cin>>aa;(p).push_back(aa);
#define vecps(p) string aa;cin>>aa;(p).push_back(aa);
#define vecp2(p) int ab;cin>>ab;(p).push_back(ab);
#define vecpl2(p) long long ab;cin>>ab;(p).push_back(ab);
#define vecps2(p) string ab;cin>>ab;(p).push_back(ab);
#define set0(k,n) for(int nn=0;nn<(n);nn++){ (k).push_back(0); }
#define sorts(c) sort((c).begin(),(c).end());
#define reverses(c) reverse((c).begin(),(c).end());
#define vec(b) vector<int> (b);
#define vecl(b) vector<long long> (b);
#define vecs(b) vector<string> (b);
#define vecsize(b,size) vector<int> (b)((size));
#define pb(b,a) (b).push_back((a));
#define doublece(a,b) cout<<(a)<<' '<<(b)<<endl;
#define pairs(s) vector<pair<int,int>> (s);
#define pairsl(s) vector<pair<ll,ll>> (s);
#define pairss(s) vector<pair<string,string>> (s);
#define pairsp(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairspl(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsps(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsREV(s) (s).push_back(make_pair(bb,aa));
#define pairslREV(s) (s).push_back(make_pair(bb,aa));
#define pairssREV(s) (s).push_back(make_pair(bb,aa));
#define MOD 1000000007

int main() {
	
string N;

cin>>N;

string ans{N.at(0)};
fx(N.size()-1){
	ans+='9';
}
ll ans2=stoll(ans.c_str());
ll ans3;
  if(ans==N){
    ans3=1;
    }else{
      ans3=0;
    }

while(ans2>0){
	ans3+=ans2%10;
	ans2/=10;
}
  
ce(ans3-1)

}