Incident tokenizes parts of the code that are repeated exactly three times (there are a few more conditions but thats the simple version). It then executes those tokens, the way it is currently structured most important token in the middle token, that is if there are n tokens the (n+1)//2 th token. We want that token to be 0mo, if it is there is a pretty good chance it will work.
If you want to be absolutely certain it will work (without testing it) you can get the same sequence of tokens as the last answer, since the last answer worked your answer will work.
You can get rid of tokens by 1) adding new copies of existing tokens 2) changing one of the instances of a token 3) causing two would be tokens to overlap
You can add new tokens by reversing any of those three