#include<bits/stdc++.h>
using namespace std;
int main(){
	int a,i,x=10,t=1,w=0;
	cin>>a;
	for(i=1;i<=a;i++){
		if(i>=x)x*=10,t++;
		w+=t&1;
	}
	cout<<w;
}
### Prompt

Create a solution in cpp for 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 a,i,x=10,t=1,w=0;
	cin>>a;
	for(i=1;i<=a;i++){
		if(i>=x)x*=10,t++;
		w+=t&1;
	}
	cout<<w;
}
```