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>
using namespace std;

#define int long long
#define fi first
#define se second
#define otp(n) cout << n;
#define ct(n) cout << n << endl;
#define cts(n) cout << n << " ";
#define cte cout << endl;
#define ctv(v) rep(i,(v).size()) ct(v[i])
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define repo(i,o,n) for(int i=o;i<n;i++)
#define repm(i,n) for(int i=n-1;i>=0;i--)
#define repv(i,v) for(int i=0;i<(v).size();i++)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define sperase(v,n) (v).erase(remove(all(v), n), (v).end());
#define vdelete(v) (v).erase(unique(all(v)), (v).end());
#define pb(n) push_back(n);

int n,f,ans,cnt,last;

main() {
    cin >> n;
    while(n) {
        ans += last = n%10;
        if (n%10 != 9 && n/10) f++;
        n/=10;
        cnt++;
    }
    if (f) ct(last+(cnt-1)*9-1)
    else ct(ans)
}