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

int main() {
	int N, COUNT = 0;;
	cin >> N;
	for (int i = 1; i <= N; i++) {
		int X = log10(i);
		if (X % 2 == 0) COUNT++;
	}
	cout << COUNT << endl;
}

### 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 N, COUNT = 0;;
	cin >> N;
	for (int i = 1; i <= N; i++) {
		int X = log10(i);
		if (X % 2 == 0) COUNT++;
	}
	cout << COUNT << endl;
}

```