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 sanyes ios_base::sync_with_stdio(0), cout.tie(0); cin.tie(0);
#define llong long long int
#define pb push_back
#define bpc __builtin_popcount
#define f first
#define s second

const llong mxn = 1e5 + 7;
const llong mod = 1e9 + 7;
const llong inf = 1e18 + 9;

using namespace std;

llong sum1 = 0, sum2 = 0;
string s;

int main(){
    cin >> s;
    sum1 = s[0] - '0' - 1;
    for (int i = 1; i < s.size(); i ++){
        sum1 += 9;
    }for (int i = 0; i < s.size(); i ++){
        sum2 += (s[i] - '0');
    }llong mx = max(sum1, sum2);
    cout << mx;
    return 0;
}