Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#pragma GCC target ("avx")
#include <bits/stdc++.h>

constexpr int INF = 2147483647;
constexpr long long int INF_LL = 9223372036854775807;
constexpr int MOD = 1000000007;
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;


int main() {
	int N;
	cin >> N;
	int ans = 0;
	for (int i = 1; i <= N; i++) {
		string a = to_string(i);
		if (a.length() % 2 == 1)ans++;
	}
	cout << ans << endl;
}