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: Runtime Error)

N=list(input())
ans=0
for i in range(len(N)-1,-1,-1):
    if i==0:
        if len(N)%2==1:
            del N[0]
            s=""
            for i in N:
                s=s+i
            s=int(s)
            s+=1
            ans+=s
    else:
        if i==len(N)-1:
            ans+=9
        elif i%2==0:
            ans+=(10**(i+1)-10**(i))

print(ans)