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 <iostream>
# include <algorithm>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <bitset>
# include <complex>
#include<limits.h>
#include<unordered_map>
#include<unordered_set>
#include<deque>
#include<cstdio>
using namespace std;
typedef long long int ll;
const int N = 1000000;
const int INF = 1 << 30;
#define rep(i,n) for(int i=(ll)0;i<(ll)n;++i)
#define ALL(x) x.begin(),x.end()
#define pp pair<ll,ll>
string YN(bool b) { return(b ? "YES" : "NO"); }
string yn(bool b) { return(b ? "Yes" : "No"); }
ll h, w, v[110][110],ct=0;
int main(){
	ll n,n2, ct = 1;
	cin >> n;
	n2 = n;
	while (n2 >= 10){++ct; n2 /= 10;}
	if (n % (ll)pow((ll)10, ct - 1) == pow(10, ct - 1) - 1)cout << 9 * (ct - 1) + n / (ll)pow((ll)10, ct - 1) << endl;
	else cout << 9 * (ct - 1) + n / (ll)pow((ll)10, ct - 1)-1 << endl;
	return 0;
}