#include<bits/stdc++.h>
using namespace std;
int main(){
	int a,b=0;
	cin >> a ;
	for(int i=a;i>=1;i--){
		int t=i,j=1;
		while(t>=10){
			t=t/10;
			j++;
		}
		if(j%2==1)
			b++;
	}
	cout << b;
	return 0;
}
### 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 main(){
	int a,b=0;
	cin >> a ;
	for(int i=a;i>=1;i--){
		int t=i,j=1;
		while(t>=10){
			t=t/10;
			j++;
		}
		if(j%2==1)
			b++;
	}
	cout << b;
	return 0;
}
```