Define the *Fouriest Transform* of a number \(n\) to be the base-\(b\) representation of \(n\) whose digits contain the most \(4\)'s of all representations such that \(b\) is minimized. If you wrote out the Fouriest Transform of all positive integers up to and including \(9952\), let \(N\) be the total number of \(4\)'s you would count in all of their digits. What are the *last three digits* of \(N\)?

**Details and Assumptions**

As an explicit example, the Fouriest Transform of \(224\) is \(1344\) (base-\(5\)) which contains \(2\) fours. Although its base-\(7\) representation, \(440\), also contains \(2\) fours, \(5\) is the smaller base.

If there does not exists a base \(b\) that \(n\) can be represented in that has any fours, define the Fouriest Transform of that number \(n\) to be its base-\(10\) representation.

This problem was inspired by the following Saturday Morning Breakfast Cereal Cartoon.

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