Hailing from a long lost tribe of vastaya, Neeko can blend into any crowd by borrowing the appearances of others, even absorbing something of their emotional state to tell friend from foe in an instant. No one is ever sure where — or who — Neeko might be, but those who intend to do her harm will soon witness her true colors revealed, and feel the full power of her primordial spirit magic unleashed upon them.

Source: https://bilibili.com/BV1ub421e7cd/

Neeko has infiltrated a grid of characters with dimensions $$$n \times m$$$, each cell in the grid is represented by a string of length $$$k$$$. She can mimic the other cells; however, Neeko's cell has at least one character different from the other cells while all the other cells are the same. Your task is to find Neeko.

You are given some special Tarot cards from the Major Arcana, including The Lovers, Death, The Star, The Moon, The Sun, and The World, along with some other playing cards in hand. You have at most one of each kind of Tarot card, and each Tarot card has a unique ability that can alter the suits of other playing cards in your hand. Your task is to determine the maximum number of flushes that can be played by the playing cards in hand after using the given Tarot cards each at most once.

The playing cards are standard playing cards, shown as below.

Here is a list of Tarot cards abilities.

• The Star: Converts up to 3 selected playing cards to Diamonds ($$$\diamondsuit$$$).
• The Moon: Converts up to 3 selected playing cards to Clubs ($$$\clubsuit$$$).
• The Sun: Converts up to 3 selected playing cards to Hearts ($$$\heartsuit$$$).
• The World: Converts up to 3 selected playing cards to Spades ($$$\spadesuit$$$).

• The Lovers: Converts 1 selected playing card into a Wild Card ($$$\circledast$$$, can be used as any suit).
• Death: Select exactly 2 playing cards, replace one card with the copy of the other card (copy the suit, the rank, and the Wild Card status).

You can use the Tarot cards in any order. Each given Tarot card can be used at most once and can never be used at all. There is a special rule about The Lovers:

• Once a playing card has been converted into a Wild Card (using The Lovers or Death), it remains a Wild Card even after applying The Star, The Moon, The Sun, and The World;
• However, if Death is used to replace a Wild Card with a copy of another card that is not a Wild Card, the resulting card will not be a Wild Card.

A flush is a set of $$$5$$$ playing cards, which could be considered a same suit: there is a suit (among Diamond, Club, Heart, and Spade) such that each of the $$$5$$$ playing cards is either of this suit or is a Wild Card.

Playing a flush means that the $$$5$$$ cards forming the flush are discarded from hand, all of which cannot be used in another flush. To the contrary of many card games including Balatro, you draw no new cards from the deck after playing cards.

Note: This is the inverse version of problem "The Emperor" with differences in constraints.

Capoo invented an interesting language named Push-Pop. This language is an interpreted language. The interpreter starts with an empty stack with infinite capacity and reads the first instruction of the custom program. There are only two kinds of instructions in this language:

• POP a GOTO x; PUSH b GOTO y

  If the top element of the stack is $$$a$$$, then pop the stack once and transfer the control flow to the $$$x$$$-th instruction (which means the next instruction will be the $$$x$$$-th). Otherwise, push an element $$$b$$$ into the stack and transfer the control flow to the $$$y$$$-th instruction.

• HALT; PUSH b GOTO y

  If the stack is empty, halt the whole program after executing this instruction. Otherwise, push an element $$$b$$$ into the stack and transfer the control flow to the $$$y$$$-th instruction.

Capoo wants to construct a Push-Pop program that halts after executing exactly $$$k$$$ instructions. Due to the naive implementation of the interpreter, a program can contain at most $$$64$$$ instructions.

Note: This is the inverse version of problem "The Empress" with differences in constraints.

Capoo invented an interesting language named Push-Pop. This language is an interpreted language. The interpreter starts with an empty stack with infinite capacity and reads the first instruction of the custom program. There are only two kinds of instructions in this language:

• POP a GOTO x; PUSH b GOTO y

  If the top element of the stack is $$$a$$$, then pop the stack once and transfer the control flow to the $$$x$$$-th instruction (which means the next instruction will be the $$$x$$$-th). Otherwise, push an element $$$b$$$ into the stack and transfer the control flow to the $$$y$$$-th instruction.

• HALT; PUSH b GOTO y

  If the stack is empty, halt the whole program after executing this instruction. Otherwise, push an element $$$b$$$ into the stack and transfer the control flow to the $$$y$$$-th instruction.

Capoo wants to upgrade the naive interpreter to deal with more instructions. Given a program of at most $$$1024$$$ instructions, calculate the number of steps the program would execute before halting.

In the ever-changing and chaotic environment of Night City, uncertainty is the only certainty. For safe travel, edgerunners rely on the best technology. Despite their premium price and corporation background, Delamain AI armed cabs provide the most reliable mode of transportation.

Today, you need to make a lengthy journey across Night City. The Delamain cab's fare structure is as follows:

• The base fare is $$$ A $$$ eurodollars, covering the first $$$ X $$$ meters.
• For the next $$$ Y $$$ meters, the fare is $$$ B $$$ eurodollars per meter.
• Beyond $$$ X + Y $$$ meters, the fare is $$$ C $$$ eurodollars per meter.

At any point, you may perform the following operation: stop and re-hail a cab. By doing so, you will immediately settle the cost of your previous cab, and the fare for the next one will be recalculated starting from the base fare. This operation can be performed for any number of times.

Now, you want to determine the minimum cost in eurodollars for traveling $$$D$$$ meters using Delamain cabs.

The renowned alchemist Xu Dog discovered that precisely removing impurities could enhance the spiritual essence of the elixirs he was refining. Through day after day of alchemy, he found that the nature of these impurities was intricately related to mathematical problems. Since your progress in Dao is still shallow, Xu Dog decided to tell you the mathematical problem he needs to solve in the most straightforward way, rather than through the esoteric problems of alchemy.

Given two positive integers $$$n \leq m$$$, calculate the sum of the answers to the following problem for all subsets of size $$$n$$$ of $$$\{1, 2, \dots, m\}$$$, modulo $$$998\,244\,353$$$:

• In a set of $$$n$$$ numbers, you can remove some numbers so that the minimum value of the set is not equal to the greatest common divisor of the set. Find the maximum number of elements that can remain in the set after removal. If no non-empty subset satisfies the condition, the answer is defined as $$$0$$$.

The greatest common divisor of a set is defined as the largest value among the common divisors of all elements in the set. For example, the greatest common divisor of the set $$$\{6, 9, 15\}$$$ is $$$3$$$.

The Hierophant and The High Priestess are fascinated by the mysteries of the universe. To explore some of these mysteries on Earth — apparently a part of the universe, they used a strange wheel to study the effect of gravity on random processes.

The wheel is a convex polygon, divided into several triangular regions by lines connecting the center of rotation to each vertex, with each triangular region representing a prize. After spinning the wheel, the prize is determined by the region directly below the stopping position. An unbiased wheel has its center of rotation at its centroid, making the final winning position dependent solely on the angle occupied by each region. Unfortunately, the wheel here may be biased: when the center of rotation is not at the centroid, the result of the wheel is always directed from the center of rotation to the centroid.

Clearly, a biased wheel is boring. To make various wheels meaningful, The Hierophant decided to perturb the result in the following way: placing a small magnetic weight at a uniformly random position within the wheel's area (considered as a point mass located inside the convex polygon) will change the centroid, thus affecting the result.

The first sample: the case without a magnet, and a possible case where the magnet changes the result.

The mass of the magnet is $$$w$$$, and the mass per unit area of the wheel is uniformly $$$1$$$. The High Priestess wants to know the probability that each region becomes the final winning region after the aforementioned perturbation.

It can be shown that the set of configurations where the wheel fails to uniquely determine the winning region (e.g., the center of rotation coincides with the centroid or lies on the boundary) has measure $$$0$$$, so you can safely ignore such cases.

People often use approximate methods to express numerical values in everyday conversations. Whether discussing time, money, or other numbers, people tend to round off to make communication simpler and easier. For example, if you and your friends dine at a restaurant with a bill of $$$98$$$ yuan, many would say, "The bill is a hundred yuan," instead of using the exact number.

If we take a more aggressive approach and round multiple times, the final result can become absurd. For instance, you could round $$$145$$$ up to $$$200$$$, because $$$145$$$ can be rounded to $$$150$$$, which can then be rounded to $$$200$$$; when someone says $$$2000$$$, it could actually have been $$$2001$$$, $$$1999$$$, $$$1888$$$, or even $$$11451$$$ before rounding.

Given a number $$$x$$$, calculate the uncertainty of $$$x$$$ within the range $$$[0,z]$$$, which is the count of numbers within the range $$$[0,z]$$$ that can be $$$x$$$ after aggressive rounding. Here, aggressive rounding is defined as performing the following rounding operation arbitrarily (possibly zero) times:

• Let the decimal representation of $$$x$$$ be $$$\overline{x_k x_{k - 1} \dots x_1}$$$, and choose an index $$$i \in \{1, 2, \dots, k\}$$$.
• If $$$x_i \lt 5$$$, subtract $$$x_i \cdot 10^{i-1}$$$ from $$$x$$$;
• otherwise, add $$$(10 - x_i) \cdot 10^{i-1}$$$ to $$$x$$$.

The rose can be viewed as a tree with $$$n$$$ nodes. To make the rose thornless, Claudette Morel can add several edges to the graph, as long as the addition does not create multiple edges or self-loops. However, she cannot add new nodes to the graph.

A simple graph is thornless if and only if each edge appears in exactly one simple cycle. A simple cycle is defined as a cycle that does not contain any repeated nodes (except for the starting and ending node being the same). The following illustrations explain what is a thornless graph and what is not.

• Left: This is a thornless graph.
• Middle: This is not a thornless graph because the edge $$$(1,2)$$$ does not appear in any simple cycle.
• Right: This is not a thornless graph because the edge $$$(1,2)$$$ appears in both cycles $$$1\sim 2\sim 5 \sim 4 \sim 1$$$ and $$$1\sim 2 \sim 5 \sim 3 \sim 1$$$.

Now, Claudette Morel has taken out her roses, and you are tasked with analyzing whether they can be transformed into a thornless graph.

The renowned farmer Chen Loong discovered that a rational planting density could raise production.

Now, the farm can be viewed as a three-dimensional coordinate system, and a plant can be seen as a point in it. There are $$$n$$$ different plants $$$A_i=(x_i,y_i,z_i)$$$. For each plant $$$A_i$$$, its density is defined as follows.

• Suppose there are $$$a$$$, $$$b$$$, and $$$c$$$ plants other than $$$A_i$$$ with the same $$$x$$$, $$$y$$$, or $$$z$$$ coordinates as $$$A_i$$$, respectively. Then, the density of $$$A_i$$$ is $$$\max\{a,b,c\}$$$.

Since Chen Loong's plants love involution, he decides to remove some plants with less density. Please answer the minimum number of plants that need to be removed such that each of the remaining plants has a density greater than or equal to $$$k$$$. Note that after removing a point, the density of other plants may change. In particular, removing all plants is always considered valid.

You need to solve for $$$k=0,1,\ldots,n-1$$$ respectively.

People use abbreviations to express succinctly. This can lead to trouble when two phrases share the same abbreviation. For example, searching the term "CCPC" online, you will find not only the desired "China Collegiate Programming Contest", but also "China Car Performance Challenge", "Competition and Consumer Protection Commission", etc.

However, nobody will drive a car while writing code. You decide to resolve the problem by assigning a distinct initialism-like abbreviation to every phrase you often use. There are $$$n$$$ phrases, each consisting of some words in lowercase and uppercase English letters. To create an abbreviation for a phrase, a non-empty prefix of each word in the phrase is chosen and then concatenated in order. For example, "ChCoPrCo" and "CCPContest" are valid abbreviations for "China Collegiate Programming Contest", but "CCCP" and "CCPiC" are not. Each phrase is abbreviated independently from the other phrases: the same word can be abbreviated differently in different phrases.

Construct an abbreviation scheme so that the total length of all abbreviations is minimized, while ensuring $$$n$$$ abbreviations are distinct.

There is a useful feature on Mikufans that allows users to leave a message during the video playback, which is called danmaku. Sometimes, there are so many danmaku messages at the same time that Kaguya cannot take them all in.

Source: https://bilibili.com/video/BV1xx411c79H

For simplicity, we only focus on the top danmaku messages: top danmaku messages are displayed at the top of the video screen, and each message occupies exactly one line. There is no limit to the number of danmaku messages at the same time (although Kaguya's screen will only display the first $$$10^9$$$ lines, the remaining messages will still be correctly maintained in the overflow area of the screen).

During the video playback, there may be three types of events:

1. A new user sends some top danmaku messages. Each message will be placed at the topmost empty line in order.
2. The danmaku messages from a specific user disappear, and the lines they are in become empty lines. The other messages will not be affected and still remain in their positions.
3. Kaguya is interested in a danmaku message, so she wants to know the sender of the top danmaku message at a specific line.

Kaguya has many videos to watch every day, and she is too busy to re-watch the video from the beginning, so she asks you for help. Please help her find the senders of the danmaku messages.

The game is played on an $$$n\times m$$$ table. A cell has its color, which is one of red, blue, or white. Initially, cell $$$(a,b)$$$ is red, cell $$$(c,d)$$$ is blue (These two cells do not coincide), and others are white. We call $$$(a,b)$$$ and $$$(c,d)$$$ special cells and others nonspecial. During the game, the player can perform a certain operation, which consists of three steps:

1. The player selects a nonspecial cell $$$(x, y)$$$.
2. The player selects another cell $$$(x', y')$$$. It must be non-white, and it must be neighboring to $$$(x, y)$$$ (i.e., $$$(x', y')$$$ and $$$(x, y)$$$ must have a common edge).
3. The cell $$$(x, y)$$$ is painted with the color of the cell $$$(x', y')$$$.

In other words, in one operation, the player can color a nonspecial cell with the same color as its non-white neighboring cell. Note that a cell may be colored more than once, and the latest color will cover the earlier one.

The player can perform the operation any number of times and then stop the game. After that, the final configuration of the map table is printed.

Unfortunately, Generalissimo is full of cheats, and cheaters can color in any position at any time. In order to advocate Justice, you decide to write a judge program to determine whether the given configuration is possible to be a legal configuration in a normal game, or there must be a cheater.