#include<stdio.h>

int main(){
  int n;
  scanf("%d", &n);
  int x=10,res=0;
  while(x<=n){
    res += x-x/10;
    x*=100;
  }
  if(n>=x/10) res += n-x/10+1;
  printf("%d\n", res);
  return 0;
}
### Prompt

Please create a solution in Cpp to 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<stdio.h>

int main(){
  int n;
  scanf("%d", &n);
  int x=10,res=0;
  while(x<=n){
    res += x-x/10;
    x*=100;
  }
  if(n>=x/10) res += n-x/10+1;
  printf("%d\n", res);
  return 0;
}
```