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 Python (3.4.3) (judge verdict: Wrong Answer)

N = int(input())

cnt = 0
i = 0
for i in range(0, 5, 2):
  m = 10 ** i
  print(m)
  if len(str(m)) > len(str(N)):
    break
  M = int("9" * (i + 1))
  print(M)
  
  if M > N:
    M = N
  print(M)
  
  cnt += M - m + 1

print(cnt)