Hello! This is day 4 of the 17 days of the Cairo Challenge. And I have no idea how to solve the playground exercise “Variables”. Can you launch this code with the instructions and write down how many steps the program has in the output?

[BONUS]: try to make it faster

// The following code outputs the 9th power of 2.
// 1. Click on "Debug" to run the code, and then click on
//    the arrows in the memory panel to trace the program
//    step by step.
//
// Observe that the computation of `z` take 8 steps and
// creates 8 temporary variables. This is because `let`
// simply replaces each instance of `y` with `x * x * x`
// and each instance of `z` with `y * y * y`, which is
// expanded to nine multiplications.
//
// 2. Change the two `let`s to `tempvar`s and run the program.
//
// "tempvar" stands for "temporary variable", which means that
// a variable will be allocated for the intermediate values
// (`x * x * x` and `y * y * y`). This reduces the same
// computation to use only 4 steps and 2 new variables.
//
// 3. Replace all three instance of the keyword `tempvar`
//    with `local`, uncomment the line containing `alloc_locals`
//    and run the code again.
//
// The behavior of the code doesn't change. This is because
// the keyword `local` allocates a new variables for the
// intermediate values as well. The main difference between
// `local` and `tempvar` is explained in the next challenge.

// Use the output builtin.
%builtins output

// Import the serialize_word() function.
from starkware.cairo.common.serialize import serialize_word

// Outputs the value 2^9.
func main{output_ptr: felt*}() {
    // alloc_locals;
    tempvar x = 2;
    let y = x * x * x;
    let z = y * y * y;
    // Because of the `let`s, the following line is equivalent to
    // serialize_word(x * x * x * x * x * x * x * x * x).
    serialize_word(z);
    return ();
}

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