#include<bits/stdc++.h>
using namespace std;
int a[100005];
int main(){
	int n,ans=0;
	scanf("%d",&n);
	for(int i=1;i<=n;++i)a[i]=a[i/10]+1;
	for(int i=1;i<=n;++i)if(a[i]&1)ans++;
	printf("%d",ans);
}
### Prompt

In cpp, your task is to solve 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 a[100005];
int main(){
	int n,ans=0;
	scanf("%d",&n);
	for(int i=1;i<=n;++i)a[i]=a[i/10]+1;
	for(int i=1;i<=n;++i)if(a[i]&1)ans++;
	printf("%d",ans);
}
```