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.

Examples

Input

11


Output

9


Input

136


Output

46


Input

100000


Output

90909
#include <bits/stdc++.h>
using namespace std;

int main(){
    int n, count=0;
    cin >> n;
    for(int i=1;i<n+1;i++){
        if(int(to_string(i).size())%2 != 0) count++;
    }
    cout << count << '\n';
}