#include<bits/stdc++.h>
using namespace std;
int f(int x)
{int r=0;
while(x>0)
{r++;
x/=10;
}
return r;
}
int main() {
int n,i,ans=0;
cin>>n;
for(i=1;i<=n;i++)
if(f(i)&1)ans++;
	cout<<ans;
}
### Prompt

Please provide a CPP coded solution 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 f(int x)
{int r=0;
while(x>0)
{r++;
x/=10;
}
return r;
}
int main() {
int n,i,ans=0;
cin>>n;
for(i=1;i<=n;i++)
if(f(i)&1)ans++;
	cout<<ans;
}
```