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

int main(){
  int N;
  cin >> N;
  int64_t ans=0;
  for(int i=1;i<=N;i++){
    if(to_string(i).size()%2!=0){
      ans++;
    }
  }
  cout << ans << endl;
}

### Prompt

In CPP, your task is to solve the following problem:

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;
  int64_t ans=0;
  for(int i=1;i<=N;i++){
    if(to_string(i).size()%2!=0){
      ans++;
    }
  }
  cout << ans << endl;
}

```