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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io;

fn digits_sum(v: i64) -> i64 {
    v.to_string().chars().map(|x| x.to_digit(10).unwrap() as i64).sum()
}

fn main() {
    let stdin = io::stdin();
    let mut buf = String::new();

    stdin.read_line(&mut buf).unwrap();

    let n: i64 = buf.trim().parse().unwrap();
    let len: u32 = buf.len() as u32;
    let head: i64 = buf[..1].parse().unwrap();

    if n < 10 {
        println!("{}", n);
    } else {
        let x = head * 10i64.pow(len-2)-1;
        println!("{}", digits_sum(x));
    }
}