#include "bits/stdc++.h"
using namespace std;
int n,ans;

int f(int x)
{
	int cnt=0;
	while(x) cnt++,x/=10;
	return cnt;
}

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		if(f(i)&1) ans++;
	cout<<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 n,ans;

int f(int x)
{
	int cnt=0;
	while(x) cnt++,x/=10;
	return cnt;
}

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		if(f(i)&1) ans++;
	cout<<ans;
}
```