Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

Solution in PHP (5.6.30) (judge verdict: Accepted)

<?php
    fscanf(STDIN, '%d', $d);

    
    if($d >=1 && $d<=9){
        $ans = $d;
    }elseif($d >=10 && $d<=99){
        $ans = 9;
    }elseif($d >=100 && $d<=999){
        $ans = 9 + ($d-99);
    }elseif($d >=1000 && $d<=9999){
        $ans = 9 + (999-99);
    }elseif($d >=10000 && $d<=99999){
        $ans = 9 + (999-99)+($d-9999);
    }else{
        $ans = 9 +(99999-9999)+(999-99);;
    }
    echo $ans;


?>