Cipher text: $4]*3\{,1,$:$\_6$#$3\_[$#$4$#$7$$$\_$#$8$#$3$#$38\_$#$8$#$1$#$7$#$7$#$3\_$#$8\_$#$7($#$3\_$#$4$#$2$#$4$#$79$#$1$#$4\}$

To decode the flag you will need to reverse three levels of encryption.

$1)$ My last step is your first. You see the cipher text that results. Unique character set, no$?$ The only problem is I had to call out which characters were actual numbers and which weren't. Then there's those pesky delimiters. That's the only clue you'll get, except that you should observe that the stories told are all about money$-$seekers, not about money$-$finders.

$2)$ We cross paths on the second step. It's funny how things stick in our heads sometimes. When I encrypted this step, I recalled my friend Vinnie from high school. I still remember his number after all these years $-742-7264.$ A real king, that guy.

$3)$ My first step is your last. Another polyalphabetic substitution cipher. I learned this technique while studying under Dr$.$ Bronckhorst at Gronsfeld. He liked to post little puzzles for students to consider on the blackboard but would never actually explain how to solve them. Apparently he never realized it's not enough to just look at a problem, you have to say a thing or two about it as well. Unless you can get two for one $($or even three!$),$ most things in life come better in pairs. Consider the key to this last step, for example, which is encoded to hide the real number values in the key: $13211321322119$

HINT: Wow! This challenge sure is proving to be a major challenge in the $2023$ contest! Or should I say Pointer Overflow $0033?$