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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
)

func main() {
	N := ReadString()
	ok := true
	for i := 1; i < len(N); i++ {
		if N[i] != '9' {
			ok = false
			break
		}
	}

	if ok {
		ans := 0
		for i := 0; i < len(N); i++ {
			ans += int(N[i] - '0')
		}
		fmt.Println(ans)
		return
	}

	if N[0] == '1' {
		fmt.Println((len(N) - 1) * 9)
		return
	}

	ans := int(N[0]-'0') - 1
	ans += (len(N) - 1) * 9
	fmt.Println(ans)
}

var reader = bufio.NewReader(os.Stdin)

func Scan(a ...interface{}) {
	if _, err := fmt.Fscan(reader, a...); err != nil {
		panic(err)
	}
}
func ReadInt() (i int)       { Scan(&i); return }
func ReadString() (s string) { Scan(&s); return }
func ReadInts(n int) []int {
	v := make([]int, n)
	for i := 0; i < n; i++ {
		Scan(&v[i])
	}
	return v
}