Customer support at Booking.com is organised as a strict reporting tree. The head of support sits at the top, and every other team member reports to exactly one person directly above them, while each person has at most two direct reports. Every team member also has a team code, a number that the quality assurance team likes to record.

Occasionally, a quality review picks three support cases handled by three different team members at once, and wants to know the single person lowest in the hierarchy who still sits on the reporting chain above every one of the three, so that one reviewer can oversee all three cases together. A team member counts as being on their own reporting chain, so if one of the three is the manager of the others, that manager is the review point. This shared reviewer is the natural escalation point that all three cases could reach by following the reporting chain upward.

You are given the layout of the whole support organisation, the team code of every member, and a long list of review requests where each request names three team members. For every request, find the review point for those three members and add together the team codes found at every review point across every request into one grand total that the quality team can record.

Implement the function sumOfTripleMeetingBadges(int n, vector<vector<long long>> nodes, int q, vector<vector<long long>> queries), which returns a long long.

The live operations desk at OpenTable, a Booking Holdings company, watches a busy Friday evening unfold one event at a time. Every minute exactly one thing happens: either a diner makes a reservation at some restaurant, or a diner cancels a reservation at some restaurant. Every restaurant in the city is identified by a numeric code so the monitoring software can process the whole evening automatically.

The desk defines the demand pressure at any minute as the combined count of active reservations at the two restaurants currently holding the most active reservations at that exact moment. When only one restaurant currently has any active reservations, the pressure is that lone count by itself, and when no restaurant has any active reservations at all, the pressure is zero.

At the end of the evening, the desk wants one number, the total demand pressure, which is simply the sum of the demand pressure recorded right after each of the n minutes. You must write a function that replays the whole evening from the event log and reports this total demand pressure.

It is guaranteed that the log never asks to cancel a reservation at a restaurant whose current count of active reservations is zero.

Implement the function sumTopTwoStockPressure(int n, vector<vector<int>> actions), which returns a long long.