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)

n = int(input())

cnt = 0

for i in range(1,n+1):
  if i < 10:
    cnt += 1
    continue
  if i >= 100 and not (i >= 1000):
    cnt += 1
    continue
  if i >= 10000 and i != 100000:
    cnt += 1
    continue

print(cnt)