Fibonacci and the number of calls

Fibonacci numbers: $F_1 = F_2 = 1$, $F_k = F_{k-1} + F_{k-2}$. They are computed by this recursive function:

fib(n):
    if n <= 2: return 1
    return fib(n - 1) + fib(n - 2)

Find $F_n$ and how many times in total the function fib is called while computing fib(n) (the first call counts too).

A single integer $n$ ($1 \le n \le 30$).

A single line with two numbers: $F_n$ and the number of calls.

Increase a global counter by $1$ inside the function. For $n = 5$: fib(5) → fib(4), fib(3); fib(4) → fib(3), fib(2); each fib(3) → fib(2), fib(1). That is $9$ calls in total, and $F_5 = 5$.

The same values are computed again and again, so the number of calls grows very fast with $n$: almost $1.7$ million for $n = 30$. This is the price of naive recursion.