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
#include<bits/stdc++.h>
using namespace std;
int a,ans;
int main(){
	scanf("%d",&a);
	for(int i=1;i<=a;i++){
		int z=i,p=0;
		while(z)z/=10,p++;
		ans+=p%2!=0;
	}
	cout<<ans<<endl;
	return 0;
}