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<iostream>
#include<cmath>
using namespace std;
int main()
{
	int n,a,i,sum=0;
	cin>>n;
	for(i=1;i<=n;i++)
	{
		a=(int)(log10(i))+1;
		if(a%2==1)
			sum++;
	}
	cout<<sum<<endl;
	return 0;
}