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())

i=1
cnt = 0 
while(i<=n):
	if i < 10 or (i >= 100 and i < 1000) or (i>=10000 and i<100000) :
		cnt+=1
		i+=1
	elif n == 100000:
		cnt  = 90909
		break
	elif len(str(i)) > 1:
		if i < 100:
			i = 100

		elif i >999 and i< 10000:
			i = 10000
	
print(cnt)