#include<iostream>
using namespace std;
int main()
{
int n,c=0,i,j,a=0;
cin>>n;
for(i=n;i>0;i--){
        j=i;
while(j>0)
{
j=j/10;
a++;
}
if(a%2==1){
    c++;
}
a=0;}
cout << c;
return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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<iostream>
using namespace std;
int main()
{
int n,c=0,i,j,a=0;
cin>>n;
for(i=n;i>0;i--){
        j=i;
while(j>0)
{
j=j/10;
a++;
}
if(a%2==1){
    c++;
}
a=0;}
cout << c;
return 0;
}

```