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


int main()
{
	int n;
	cin >> n;

	int ans = 0;
	for (int i = 1; i <= n; i++) {
		if (to_string(i).size() % 2 == 1) {
			ans++;
		}
	}

	cout << ans << 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;


int main()
{
	int n;
	cin >> n;

	int ans = 0;
	for (int i = 1; i <= n; i++) {
		if (to_string(i).size() % 2 == 1) {
			ans++;
		}
	}

	cout << ans << endl;
}

```