#include<bits/stdc++.h>
using namespace std;
int w,q;
int main() {
	cin>>q;
	if(q==100000) q--;
	for(int l=1;l<=q;l++) {
		if(l>=10&&l<100||l>=1000&&l<10000) continue;
		w++;
	}
	cout<<w;
	return 0;
}
### Prompt

Generate a Cpp solution to 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 w,q;
int main() {
	cin>>q;
	if(q==100000) q--;
	for(int l=1;l<=q;l++) {
		if(l>=10&&l<100||l>=1000&&l<10000) continue;
		w++;
	}
	cout<<w;
	return 0;
}
```