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 = input()
digit_n = len(str(n))
n = int(n)
ans = 0

if digit_n % 2 == 0:
    for i in range(digit_n // 2):
        ans += 9 * 10 ** (i * 2)

else:
    if digit_n != 1:
        for i in range((digit_n - 1) // 2):
            ans += 9 * 10 ** (i * 2)
        hasuu = int(str(n)[1:])
        ans += hasuu + 1
    
    else:
        ans = n

    
print(ans)