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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> l_l;
typedef pair<int , int> i_i;
typedef vector<ll> vel;
typedef vector<int> vei;
typedef vector<char> vec;
typedef vector<bool> veb;
typedef vector<string> ves;
typedef vector<vector<ll>> ve_vel;
typedef vector<vector<int>> ve_vei;
typedef vector<vector<char>> ve_vec;
typedef vector<vector<bool>> ve_veb;
typedef vector<vector<string>> ve_ves;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<(int)(n);i++)
#define rep2(i,n) for(int i=2;i<(int)(n);i++)
#define repk(i,k,n) for(int i=k;i<(int)(n);i++)
#define fs first
#define sc second
#define pub push_back
#define puf push_front
#define pob pop_back
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define maxel(a) *max_element(all(a))
#define minel(a) *min_element(all(a))
#define acc accumulate
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define mod (1000000007)
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

    string s; cin >> s;

    if(s.size() == 1) {cout << s << endl; return 0;}
    else {
        bool check = true;
        rep1(i,s.size()) {
            if(s[i] != 9) {check = false; break;}
        }
        if(check) {cout << (s[0] - '0') + 9*(s.size()-1) << endl; return 0;}
        else {cout << (s[0]-'0') + 9*(s.size()-1)-1 << endl; return 0;}
    }
}