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: Wrong Answer)

#include<iostream>
#include<cmath>
using namespace std;

int main() {

	int64_t N, digit, a[16], i, sum = 0;
	cin >> N;
	digit = int64_t(log10(N)) + 1;

	if (digit == 1) {
		cout << N << endl;
		return 0;
	}

	for (i = 0; i < digit; ++i) {
		a[i] = N / int64_t(pow(10, digit-1));
		N = N % int64_t(pow(10, digit-1));
	}

	for (i = digit - 1; i >= 0; --i) {
		if (i == 0) --a[i];
		else if (a[i] != 9) a[i] = 9;
		else continue;
	}

	for (i = 0; i < digit; ++i) {
		sum += a[i];
	}

	cout << sum << endl;
	return 0;
}