#include<iostream>
using namespace std;

int main(){
	int a,ans=0,k,w;
	cin>>a;
	for(int i=1;i<=a;++i){
		k=i,w=1;
		while(k>9)k/=10,w++;
		if(w&1)ans++;
	}
	cout<<ans;
	return 0;
}
### 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<iostream>
using namespace std;

int main(){
	int a,ans=0,k,w;
	cin>>a;
	for(int i=1;i<=a;++i){
		k=i,w=1;
		while(k>9)k/=10,w++;
		if(w&1)ans++;
	}
	cout<<ans;
	return 0;
}
```