How the output of this code is come as 17. When I calculated it gives me 18?

Question

#include <stdio.h>
int f(int n)
{
static int r = 0;
if (n <= 0)
return 1;
if (n > 3)
{
r = n;
return f(n-1)+r;
}
return f(n-2)+r;
}
int main()
{
printf("%d", f(5));
}

Please explain the steps of code.

Answer 1

I think 18 seems like the right answer.
Function calls(working):
1.f(n-1)+r;//n=5
2.f(n-1)+r;//n=4
3.f(n-2)+r;//n=3
4.f(n-2)+r;//n=1
5.1;//n=-1.
Edit:
Now, after function call returns the final value 1, we add the returned values to the value of the static variable r(which now has value 4 for any function call).
((((1+4)+4)+4)+4)
=(((5+4)+4)+4)
=((9+4)+4)
=(13+4)
=17.
Point is evaluation is done after last function returns a value so static variable r's value remains unchanged i.e. 4.

Answer 2

Arnold D'Souza but on executing the code will return 17 so I think there is something which I'm doing wrong