Turn fee flags into stackable decorators.

A delivery app prices an order with one function and a flag per add-on: a distance fee, a small-cart fee, a late-night fee and a surge. Every new add-on is another flag, every caller changes, and the order of the rules is hidden in the order of the if statements. Rebuild it as decorators: each add-on is a small object that takes any fee rule and is itself a fee rule.

The behaviour to keep, exactly (an Order has km, cart in rupees and hour from 0 to 23):

Build this API (names are Python's; JavaScript and TypeScript use the same names):

• DistanceFee(), with fee(order).
• SmallCart(inner), LateNight(inner) and Surge(inner, times): each has fee(order), calls inner.fee(order) and then applies its own rule to the answer. They work with any object that has fee(order).
• build(names): starts from a DistanceFee and wraps once per name, in list order, so the first name sits closest to the base and the last is outermost. The names are "small-cart", "late-night" and "surge" (surge uses 2). Any other name raises ValueError (JavaScript and TypeScript: an Error) with the message unknown add-on: <name>.

The tests check every old number, that wrapping order changes the answer (210 against 175), that a decorator accepts another decorator, that a new add-on written inside a test plugs in without editing anything, that fee never changes its order or its inner rule, and that no flag is left in the code.