#include<bits/stdc++.h>
using namespace std;
int n,i,j,a,c;
string s;
int main()
{
  for(i=1,cin>>n;i<=n;i++){
    for(c=0,j=i;j!=0;j/=10)c++;
    if(c&1)a++;
  }
  cout<<a;
}
### 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<bits/stdc++.h>
using namespace std;
int n,i,j,a,c;
string s;
int main()
{
  for(i=1,cin>>n;i<=n;i++){
    for(c=0,j=i;j!=0;j/=10)c++;
    if(c&1)a++;
  }
  cout<<a;
}
```