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 JavaScript (node.js v5.12) (judge verdict: Accepted)

"use strict"
function main(arg) {
    //最大は1000000
    var num = parseInt(arg);
    var kotae = 0;


    for(var i = 1; i <= num; i++){
        var mozii =String(i);
        if(mozii.length % 2 != 0){
            kotae++;
        }
    }



    console.log(kotae);


}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));