Lois Griffin has hung a new sign in the kitchen: "But first, coffee." Peter finds this deeply illogical, as you can't say something comes "first" without knowing what it's being compared against.

To settle the argument, Peter gathers a line of $$$n$$$ people in the kitchen, represented as a string $$$s$$$. Each person either wants coffee $$$(s_i = \texttt{C})$$$ or is having a life-threatening emergency $$$(s_i = \texttt{E})$$$.

Lois insists that coffee should always come first, so she wants to rearrange the line so that all $$$\texttt{C}$$$'s appear before all $$$\texttt{E}$$$'s. Lois can only perform the following operation, any number of times:

— Choose 3 indices $$$1 \leq i \lt j \lt k \leq n$$$ and right shift or left shift the values on $$$s_i$$$, $$$s_j$$$, $$$s_k$$$ cyclically.

For the string $$$\texttt{CCECCE}$$$, if she chooses $$$i=1, j=2, k=3$$$ and performs a right shift cyclically, the string becomes $$$\texttt{ECCCCE}$$$. If she chooses $$$i=4, j=5, k=6$$$ and performs a left shift cyclically instead, the string becomes $$$\texttt{CCECEC}$$$.

Determine the minimum number of operations Lois needs to rearrange $$$s$$$ so that all people who want coffee come before all people having emergencies.

By day, Coffee Man is an ordinary office worker. But give him a cup of coffee and he becomes a superhero who can leap across the city in a single bound!

Coffee Man is standing at point $$$0$$$ of a number line, and he needs to get to point $$$n$$$ (reaching any point with coordinate at least $$$n$$$ is enough). He brought $$$m$$$ cups of coffee; the $$$i$$$-th cup initially has temperature $$$a_i$$$.

Every second, Coffee Man does exactly one of the following:

• walk: his coordinate increases by $$$1$$$;
• drink a cup he has not drunk yet: his coordinate increases by the current temperature of that cup.

Each cup can be drunk at most once. Unfortunately, coffee gets cold: after every second, the temperature of every cup decreases by $$$1$$$, but it never goes below $$$0$$$. In other words, if Coffee Man drinks the $$$i$$$-th cup when exactly $$$s$$$ seconds have passed, he moves forward by $$$\max(0, a_i - s)$$$.

What is the minimum number of seconds Coffee Man needs to reach a point with coordinate at least $$$n$$$?

It's 4 PM on a Wednesday, and Bevo just finished a 2-hour OS lecture. He signed up for the UTPC contest this evening, and needs some caffeine to keep him going. But unlike every other CS student, Bevo doesn't LinkedInMaxx. Bevo doesn't larp. Bevo doesn't network. Bevo doesn't swing by Up and Atom in Welch for matcha or buy an $8 coffee from Lucky Lab. Bevo only drinks black coffee straight from the pot that costs 2 cents to make, like a middle-aged man who has a wife and three kids at home. To do this, he must quickly head back to his apartment so he can depart for the contest on time.

Consider the GDC as location $$$1$$$ and Bevo's apartment as location $$$N$$$. UT campus has $$$N - 2$$$ street corners, numbered $$$2$$$ through $$$N - 1$$$. Each street connects two of these locations (a corner, the GDC, or the apartment) and is accompanied by an estimated time to walk across it. Streets are two-way. Help Bevo rush home in the shortest possible time so he can refuel.

Ballerina Cappuccina has opened a coffee tasting stand and prepared $$$n$$$ $$$(1 \le n \le 2 \cdot 10^5)$$$ cups of coffee, where $$$n$$$ is odd.

Each cup has an integer strength. More formally, let $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^9)$$$ be the strengths of the cups.

Before the tasting begins, Cappuccina must discard exactly one cup. She then pairs up all of the remaining cups.

The cost of pairing two cups with strengths $$$x$$$ and $$$y$$$ is $$$|x-y|$$$. The total cost is the sum of the costs of all pairs.

Cappuccina wants the flavors in each pair to be as similar as possible. Help her determine the minimum total cost she can achieve by choosing which cup to discard and how to pair the remaining cups optimally.

Johnny is a busy man – he has many events scheduled on his calendar. As Johnny has some school assignments that he hasn't completed yet, he wants to keep Claude Code running on his computer for as long as possible, but whenever he is away at an event, his computer shuts down immediately by default to save power. Fortunately, Johnny can use the command line utility caffeinate to keep his computer running. In particular, if Johnny uses the command caffeinate, his computer will stay on for $$$T$$$ minutes, regardless of whether or not Johnny uses his computer in that time window.

Johnny can use the caffeinate command up to $$$k$$$ times at any moment he chooses (even if he is away at an event). Given $$$n$$$ events on Johnny's calendar, Johnny wants to know – what is the maximum amount of time Johnny can run Claude Code for when he is away at events?

On his voyage to hunt the elusive white whale, Captain Ahab is swamped with tasks to do during day-to-day life on his ship. The good news is that he has the life-sustaining power of coffee on his side. The bad news is that there is a limited amount of coffee in his ship's hold. There are $$$n$$$ days, each with some amount of work $$$w_i$$$ to do. Each unit of work costs 1 energy, and every day starts with 0 energy.

There are $$$m$$$ cups of coffee that can be drunk on any day. The first coffee drunk on any day gives $$$k$$$ energy, the second gives $$$k-1$$$ energy, and so on, until the $$$k^{th}$$$ coffee gives $$$1$$$ energy; any coffee drunk afterward gives no energy.

Ahab needs your help to ration his coffee to complete as much work as possible.

Jihyo and Dahyun are tired from songwriting, so they decide to play a game to kill time. They usually play janggi (Korean chess), but decide to try out a new game.

They have a computer that randomly creates a directed graph: $$$n$$$ circles are drawn on the table, and $$$m$$$ arrows connect from one node (circle) to another. It is guaranteed no arrow will point from a node to itself, and it is guaranteed there won't be multiple arrows that have the same start and end node. However, note it is possible (as an example) to have an arrow point from Node 1 to Node 2, and another arrow point from Node 2 to Node 1. Furthermore, each node is either painted apricot (for Jihyo) or kept white (for Dahyun).

When they're ready to play, the computer will randomly pick a starting node for Jihyo among all of the apricot nodes, and a starting node for Dahyun among all of the white nodes. The two will then take turns moving, with Jihyo always going first. When it is a player's turn, they must pick an edge pointing out from their node and travel to where it points. A player is always allowed to re-visit a node, but they can never travel to a node that their opponent is currently on. If a player doesn't have any valid moves, the game ends and that player loses. If after $$$10^7$$$ moves (since both players are very patient) the game does not end, it ends in a draw. Both players will naturally play optimally (preferring wins to draws and draws to losses).

Dahyun is convinced that the computer could potentially be rigged against her, especially since she always has to go second. To test her theory, she wants to figure out the probability the game ends in a draw and the probability game ends in her loss. Can you help her find this out quickly?

Lucky Lab A is west of Guadalupe Street, while Lucky Lab B is east of it. There are $$$n$$$ loyal customers of Lucky Lab A waiting on the east sidewalk and $$$m$$$ loyal customers of Lucky Lab B waiting on the west sidewalk. Customer $$$i$$$ of Lucky Lab A is at vertical position $$$a_i$$$, and customer $$$i$$$ of Lucky Lab B is at vertical position $$$b_i$$$. Multiple customers may begin at the same position.

Before crossing, every customer may walk along their current side of the street to any integer vertical position. They then cross Guadalupe in a straight horizontal line. Customers traveling in the same direction may cross together, but a customer heading west and a customer heading east cannot use the same vertical position, or they would collide.

Walking from vertical position $$$x$$$ to vertical position $$$y$$$ costs $$$|x-y|$$$. Find the minimum possible total walking cost that lets everyone cross without a collision.