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: Accepted)

A, = [list(map(int, s.split())) for s in open(0)][0]

ans = 0
num_digits = len(str(A))
if num_digits == 1:
    ans += A 
else:
    if num_digits % 2 != 0:
        ans += A - 10 ** (num_digits - 1) + 1
    for i in range(3, num_digits, 2):
        ans += 10 ** i - 10 ** (i - 1) 
    ans += 9
print(int(ans))