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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define pa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
#define pb push_back

using ll = long long;
using namespace std;
using pint = pair<int,int>;
using pll = pair<ll,ll>;
const int INFint = 1e9;
const ll INFll = 1e18;
ll MOD=1e9+7;

ll gcd(ll a, ll b) { return __gcd(a,b); } //最大公約数
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; } //最大公倍数

// aのn乗をMODで割りながら計算する
ll modpow(ll a, ll n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % MOD;
    a = a * a % MOD;
    n >>= 1; // right shift
  }
  return res;
}

// MODを法としたaの逆元を計算する
ll modinv(ll a) {
  return modpow(a, MOD - 2);
}

int main(){
  ll N;
  cin>>N;
  int len = int(to_string(N).size());
  string s = to_string(N);
  bool flag(true);
  for (int i(0);i<len-1;i++){
    if (to_string(N)[i+1] != '9'){
      flag = false;
    }
  }
  int cnt = int(to_string(N).size()) - 1;
  if (flag) cout << 9*cnt+(to_string(N)[0]-'0') << endl;
  else{
    cout << cnt*9 + (to_string(N)[0]-'1') << endl;
  }
  return 0;
}