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())

k=str(n)
ans=0

for i in range(1,len(k),2):
  s="9"
  s*=i
  ans+=int(s)-int(s)//10
    
if(len(k)%2==1):
	m=pow(10,len(k)-1)
	ans+=n-m+1

print(ans)