Imagine that you have 2 eggs and are standing at the bottom of a very large skyscraper. Let’s say that skyscraper has `n` floors.

There is a certain floor, `f`, where if you drop an egg from that height (or a height above that floor), the egg will break. If you drop an egg below floor `f`, the egg won’t break.

To find this floor `f`, you decide to start dropping your eggs from different floors. If you drop an egg from a floor, and that egg breaks, you can no longer use that egg.

To find floor `f`, one strategy you could use would be to start dropping your egg from the bottom of the skyscraper, and keep going up a floor one at a time until the egg breaks. But that sounds like a lot of work dropping all of those eggs! You want to know what the minimum number of drops it would take to find floor `f`.

Complete the function `egg_drop(n)` that returns the minimum number of egg drops it would take to discover what floor of your tower is floor `f`, given that the tower has `n` total floors. Remember, don’t return floor `f` — return the number of egg drops it would take to find floor `f`.

For example `egg_drop(2)` should return 2. If the tower has `2` stories, then we need to commit to dropping an egg twice. We could drop an egg from the first floor. If the egg breaks, we got lucky, and now know that `f = 1`. But if the egg doesn’t break, then we need to drop an egg from floor `2` to see if that egg breaks. If it does, then we know `f = 2`. If it doesn’t break, then we know `f` is greater than the height of the tower.

Things get trickier as `n` increases. For example, if `n` is `100`, then the minimum number of drops it will take to find `f` is `14`.

This challenge was reported to have been asked at interviews with Microsoft. If you’ve covered the material in Pass the Technical Interview with Python or an equivalent, you should be able to solve this challenge. If you have trouble, try thinking about this problem recursively. Our lesson on recursion might help!