Report Chain / Org Tree

You are given a list of employee-manager pairs representing the reporting structure of a company. Each pair is of the form [employee, manager] where both are strings. The CEO has no manager (no incoming edge). Build the organizational tree and support two queries:

1. find_chain(employee): print the chain of command from the given employee up to the CEO, one name per line with increasing indentation (two spaces per level) to show depth.
2. find_common_manager(e1, e2): return the lowest common manager (LCM) of the two employees. If one employee is an ancestor of the other, that ancestor is the LCM. If either employee does not exist, return null.

Input is first the number of pairs, followed by that many pairs. After the pairs, commands are given: "chain <employee>" or "common <e1> <e2>". Process each command and output accordingly.