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

int main() {
int N;
  cin>>N;
  int counter=0;

  for(int i=1;i<N+1;i++){
if((to_string(i)).size()%2==1){

      counter++;}
  }
  cout <<counter<<endl;
    
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

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

int main() {
int N;
  cin>>N;
  int counter=0;

  for(int i=1;i<N+1;i++){
if((to_string(i)).size()%2==1){

      counter++;}
  }
  cout <<counter<<endl;
    
}

```