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 Common Lisp (SBCL 1.1.14) (judge verdict: Accepted)

(defun digitsp (n r)
  (and (<= (expt 10 (- r 1)) n)
       (< n (expt 10 r))))

(defun odd-digitsp (n)
  (or (digitsp n 1)
      (digitsp n 3)
      (digitsp n 5)))

(let ((n (read)))
  (princ (loop :as i
               :below n
               :when (odd-digitsp (+ 1 i))
               :count i)))