#include<bits/stdc++.h>
using namespace std;

long long N;
int main(){
	cin >> N;
	int count = 0;
	for(int i = 1; i<=N; i++){
		if(to_string(i).size() % 2 == 1){
			count++;
		}
	}
	cout << count << endl;
}


### Prompt

Create a solution in Cpp for 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;

long long N;
int main(){
	cin >> N;
	int count = 0;
	for(int i = 1; i<=N; i++){
		if(to_string(i).size() % 2 == 1){
			count++;
		}
	}
	cout << count << endl;
}


```