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 PyPy3 (2.4.0) (judge verdict: Accepted)

# B
def keta(n):
    count = 0
    while n != 0:
        n = n//10
        count+=1
    return count

n  = int(input())
k = keta(n)
if k == 1:
    print(n)
elif k == 2:
    print(9)
elif k == 3:
    print(9 + n-99)
elif k == 4:
    print(909)
elif k == 5:
    print(909 + n-9999)
elif k == 6:
    print(90909)