#include<bits/stdc++.h>
using namespace std;
int main(){
  int n,cnt=0;
  cin>>n;
  for(int i=1;i<=n;++i)
    if((i>=1&&i<10)||(i>=100&&i<1000)||(i>=10000&&i<100000)) ++cnt;
  cout<<cnt<<endl;
  return 0;
 } 
### 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,cnt=0;
  cin>>n;
  for(int i=1;i<=n;++i)
    if((i>=1&&i<10)||(i>=100&&i<1000)||(i>=10000&&i<100000)) ++cnt;
  cout<<cnt<<endl;
  return 0;
 } 
```