P12996 [GCJ 2022 #2] I, O Bot

To welcome attendees to a developers' conference on Jupiter's moon of Io, the organizers inflated many giant beach balls. Each ball is in roughly the shape of either a 1 or a 0, since those look sort of like the letters I and O. The conference just ended, and so now the beach balls need to be cleaned up. Luckily, the beach ball cleanup robot, BALL-E, is on the job! The conference was held on an infinite horizontal line, with station 0 in the middle, stations 1, 2, … to the right, and stations $-1, -2, \ldots$ to the left. Station 0 contains the conference's only beach ball storage warehouse. Each other station contains at most one beach ball.  BALL-E has two storage compartments, each of which can hold a single beach ball. One compartment can only hold 1-shaped balls and the other can only hold 0-shaped balls. (The 1-shaped balls are more oblong than the 0-shaped balls, so neither shape of ball will fit in the other shape's compartment.) BALL-E initially has both the 0 and 1 compartments empty, and it starts off at station 0. The robot can do the following things: * Move left one station or right one station. This costs 1 unit of power. * If there is a ball at the current station, and BALL-E is not already storing a ball of that shape, it can put the ball in the appropriate compartment. This takes 0 units of power. * If there is a ball at the current station, BALL-E can compress it so that its shape becomes the other shape. That is, a 1-shaped ball becomes a 0-shaped ball, or vice versa. It takes $\mathbf{C}$ units of power to do this. Note that BALL-E may not change the shape of a ball that it has already put into one of its compartments. * If BALL-E is at station 0 and is storing at least one ball, it can deposit all balls from its compartment(s) into the beach ball storage warehouse. This takes 0 units of power and leaves both compartments empty. Notice that if BALL-E moves to a station and there is a ball there, BALL-E is not required to pick it up immediately, even if the robot has an open compartment for it. Also, if BALL-E moves to the station with the warehouse, it is not required to deposit any balls it has. Find the minimum number of units of power needed for BALL-E to transfer all of the balls to the warehouse, using only the moves described above.

The first line of the input gives the number of test cases, $\mathbf{T}$. $\mathbf{T}$ test cases follow. The first line of each test case contains two integers, $\mathbf{N}$ and $\mathbf{C}$: the number of balls and the amount of power units needed to change the shape of a ball, respectively. The next $\mathbf{N}$ lines describe the positions (i.e., station numbers) and the shapes of the balls. The $i$-th line contains two integers, $\mathbf{X}_{\mathbf{i}}$ and $\mathbf{S}_{\mathbf{i}}$: the position and the shape of the $i$-th ball, respectively.

For each test case, output one line containing case # $x$: $y$, where $x$ is the test case number (starting from 1) and $y$ is the minimum number of units of power needed to transfer all of the balls to the warehouse, as described above.

**Sample Explanation** In Sample Case #1 (illustrated in the statement), there are $\mathbf{N} = 5$ balls and $\mathbf{C} = 0$. One optimal strategy is to make three round trips from (and back to) the warehouse: * First round trip: Move to station 3, pick up the 0 ball there and store it in the 0 compartment, move back to station 0, and deposit the ball in the warehouse. This takes 6 units of power. * Second round trip: Move to station 8, pick up the 0 ball there, and store it in the 0 compartment. Move to station 6, change the 0 ball there into a 1 ball, and pick it up and store it in the 1 compartment. Move to station 0 and deposit both balls in the warehouse. This takes 16 units of power. (Recall that in this case, it takes 0 units of power to change a ball's shape.) * Third round trip: Move to station 10. Change the 1 ball there into a 0 ball, and pick it up and store it in the 0 compartment. Move to station 15. Pick up the 1 ball there and store it in the 1 compartment. Move to station 0 and deposit both balls in the warehouse. This takes 30 units of power. The total number of units of power needed to collect all the balls is 52. Sample Case #2 is like Sample Case #1, but now with $\mathbf{C} = 10$. Now BALL-E has to use at least 56 units of power: * First round trip: Get the ball from station 3. This takes 6 units of power. * Second round trip: Get the differently-shaped balls from stations 6 and 10. (These are a 0 and a 1, respectively, so there is no need to change the shape of either of them.) This takes 20 units of power. * Third round trip: Get the differently-shaped balls from stations 8 and 15. This takes 30 units of power. Sample Case #3 is also like Sample Case #1, but now with $\mathbf{C} = 1$. Here, BALL-E needs at least 54 units of power: * First round trip: Get the ball from station 3. This takes 6 units of power. * Second round trip: Get the ball from station 8. When passing through station 6 on the way back, change the shape of the ball there and get it. This takes 17 units of power. * Third round trip: Do the same with the balls at stations 15 and 10. This takes 31 units of power. In Sample Case #4, one optimal strategy is for BALL-E to move to station $-1000000000$, get the 1 ball there, move to station $1000000000$, get the 0 ball there, and then return to station 0 to deposit both of them. **Limits** - $1 \leq \mathbf{T} \leq 100$. - $0 \leq \mathbf{S}_{\mathbf{i}} \leq 1$, for all $i$. - $-10^{9} \leq \mathbf{X}_{\mathbf{i}} \leq 10^{9}$, for all $i$. - $0 \leq \mathbf{C} \leq 10^{9}$. - $\mathbf{X}_{\mathbf{i}} \neq 0$, for all $i$. - All $\mathbf{X}_{\mathbf{i}}$ are distinct. **Test Set 1 (11 Pts, Visible Verdict)** For at most 15 cases: - $1 \leq \mathbf{N} \leq 5000$. For the remaining cases: - $1 \leq \mathbf{N} \leq 100$. **Test Set 2 (20 Pts, Hidden Verdict)** For at most 15 cases: - $1 \leq \mathbf{N} \leq 10^{5}$. For the remaining cases: - $1 \leq \mathbf{N} \leq 5000$.