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: Wrong Answer)

#include <iostream>
#include <vector>
#include<algorithm>
#include<iomanip>
#include<queue>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for(int i=m;i<(int)(n);i++)
typedef pair<double,double> pint;
typedef pair<pint,int> p3;
typedef long long ll;
const int inf   = 1<<29;
const ll  longinf = 1LL<<60;
int dx[4]={1,0,-1,0}, dy[4]={0,1,0,-1};

int main(){
	string s;
	cin>>s;
	int ans=0;
	int n=s.size();
	rep(i,n)if(s[i]!='0'){
		int res=0;
		rep(j,i+1){
			res+=s[j]-'0';
		}
		rep(j,n-i-1){
			res+=9;
		}
		ans=max(ans,res-1);
	}
	cout<<ans<<endl;
	return 0;

}