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: Runtime Error)

#include <vector> 
#include <list> 
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <queue>
using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

//math
//-------------------------------------------
template<class T> inline T sqr(T x) { return x * x; }

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI = acos(-1.0);

//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))

//debug
#define dump(x)  cerr << #x << '=' << (x) << endl;
#define debug(x) cerr << #x << '=' << (x) << '('<<'L' << __LINE__ << ')' << ' ' << __FILE__ << endl;

#define YES_ cout<<"YES"<<endl; return 0;
#define NO_ cout<<"NO"<<endl; return 0;



int main() {

	string S;
	LL ans = 0LL;
	
	cin >> S;
	int n = S.size();

	LL jimei = 9*(S.size()-1);


	if (S[0] != '9') {
		
		//先頭が何かの数字かつ下の桁がすべて9ならその数字が答え
		FOR(i, 1, S.size()) {
			if (S[i] != '9') { goto NEXT; }
		}
		FOR(i, 0, S.size()) {
			ans += LL(S[i]-'0');
		}
		cout << max(ans,jimei) << endl;
		return 0;

	NEXT:;
		//したがすべて９ではなかった
		ans += S[0] - '1';
		REP(i, S.size() - 1) {
			ans += 9;
		}
		cout << max(ans,jimei) << endl;
		return 0;
	}
	//先頭が9だった
	else {
	
		FOR(i, 1, S.size()) {
			if (S[i] != '9') { goto NONALL; }
		}
		//全て9
		REP(i, S.size()) {
			ans += 9;
		}
		cout << max(ans, jimei) << endl;

		return ans;

	NONALL:;
		ans += 8;
		FOR(i, 1, S.size()) {
			ans += 9;
		}

		cout << max(ans,jimei) << endl;
		return 0;

	}
	



}