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, i, ans=0;
  cin >> n;
  for(i=1; i<=n; i++){
    if((to_string(i).length())%2==1){
      ans+=1;
    }
  }
  cout << ans << endl;
  return 0;
}