#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
int n;cin>>n;int count=0;
for(int i=1;i<=n;i++){
    string s=to_string(i);
    if(s.size()%2==1)count++;
}
cout<<count<<endl;
}
### Prompt

Your task is to create a Cpp solution 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;
typedef long long ll;
int main(){
int n;cin>>n;int count=0;
for(int i=1;i<=n;i++){
    string s=to_string(i);
    if(s.size()%2==1)count++;
}
cout<<count<<endl;
}
```