The Fibonacci numbers are defined as follows: $f_0 = 0$, $f_1 = 1$, and $f_n = f_{n-1} + f_{n-2}$ for
		$n \geq 2$. 
		
		Consider again the recursive algorithm $\Fib$, which takes as input an integer $n \geq 0$:
		
		
			$\mathbf{Algorithm}\ \Fib(n)\mathrm{:}$ 
			$\mathbf{if}\ n = 0\ \mathrm{or}\ n = 1$ 
			$\mathbf{then}\ f = n$ 
			$\mathbf{else}\ f = \Fib(n - 1) + \Fib(n - 2)$ 
			$\mathbf{endif};$ 
			$\mathbf{return}\ f$ 
		
		
		For $n \geq 3$, run algorithm $\Fib(n)$ and let $a_n$ be the number of times that $\Fib(2)$ is called. 
		Which of the following is true?
    
(a)
 For $n \geq 3$, $a_n = -1 + f_n$
   
(b)
 For $n \geq 3$, $a_n = f_n$
   
(c)
 For $n \geq 3$, $a_n = f_{n + 1}$
   
(d)
 For $n \geq 3$, $a_n = f_{n - 1}$