 The last thing we need to do is show the A transitions out of states 0, 1, and 2. Let's take another step back from our automaton and think about it logically. We have any number of B's in a row, and then we get an A. That breaks our row of B's, and we go back to having 0 B's in a row. Now back to our automaton. We have 0, 1, or 2 B's in a row, and then we get an A. We just decided that we go back to the beginning, so let's do it. Our A transitions are from 0 to 0, 1 to 0, and 2 to 0. So have we finished? Let's check. State 0 has an A transition and a B transition. State 1 has an A transition and a B transition. State 2 has an A transition and a B transition. And State 3 has an A transition and a B transition. We have a start state, 0, and we have our accepting states, 0, 1, and 2. Awesome! All we have to do now is make sure it works. I'll leave you to run through it with a few test cases. If you're watching on YouTube, then leave us a comment. Let us know how you go, or if you have any feedback on the series of videos. And for more information and examples, check out the Computer Science Field Guide. Follow the link in the description below and navigate to Chapter 14, Conformal Languages. Thanks for watching!