Problem:
Score: 500 points

Problem Statement

New AtCoder City has an infinite grid of streets, as follows:

At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.

A straight street, which we will call East-West Main Street, runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.

A straight street, which we will call North-South Main Street, runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.

There are N residential areas in New AtCoder City. The i-th area is located at the intersection with the coordinates (X_i, Y_i) and has a population of P_i. Each citizen in the city lives in one of these areas.

The city currently has only two railroads, stretching infinitely, one along East-West Main Street and the other along North-South Main Street.

M-kun, the mayor, thinks that they are not enough for the commuters, so he decides to choose K streets and build a railroad stretching infinitely along each of those streets.

Let the walking distance of each citizen be the distance from his/her residential area to the nearest railroad.

M-kun wants to build railroads so that the sum of the walking distances of all citizens, S, is minimized.

For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after building railroads?

Constraints

1 \leq N \leq 15

-10 \ 000 \leq X_i \leq 10 \ 000

-10 \ 000 \leq Y_i \leq 10 \ 000

1 \leq P_i \leq 1 \ 000 \ 000

The locations of the N residential areas, (X_i, Y_i), are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N

Output

Print the answer in N+1 lines.

The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value of S after building railroads for the case K = i-1.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

8
2 2 286017
3 1 262355
2 -2 213815
1 -3 224435
-2 -2 136860
-3 -1 239338
-2 2 217647
-1 3 141903

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)

import kotlin.math.abs
import kotlin.math.min

fun main() = msolutions2020e()

@OptIn(ExperimentalStdlibApi::class)
fun msolutions2020e() {
    val n = readLine()!!.toInt()
    val xList = IntArray(n)
    val yList = IntArray(n)
    val pList = IntArray(n)
    repeat(n) { i ->
        val (x, y, p) = readLine()!!.split(" ").map { it.toInt() }
        xList[i] = x
        yList[i] = y
        pList[i] = p
    }

    val choosePatterns = 1 shl n
    // 何もしてないのときの移動距離
    val baseDistances = (0 until n).map { min(abs(yList[it]), abs(xList[it])) }.toIntArray()
    // [その集合の街を基準に『横』に線路を引いたとき][i番目の街の] = 移動距離の最小
    val costsVert = Array(choosePatterns) { baseDistances.copyOf() }
    // [その集合の街を基準に『縦』に線路を引いたとき][i番目の街の] = 移動距離の最小
    val costsHorz = Array(choosePatterns) { baseDistances.copyOf() }

    val diffVert = Array(n) { IntArray(n) { Int.MAX_VALUE } }
    val diffHorz = Array(n) { IntArray(n) { Int.MAX_VALUE } }

    for (from in 0 until n) for (to in 0 until n) {
        diffVert[from][to] = min(diffVert[from][to], abs(yList[from] - yList[to]))
        diffHorz[from][to] = min(diffHorz[from][to], abs(xList[from] - xList[to]))
    }

    for (choose in 0 until choosePatterns) {
        for (from in (0 until n)) {
            if ((choose shr from) and 1 == 0) continue
            for (to in 0 until n) {
                costsVert[choose][to] = min(costsVert[choose][to], diffVert[from][to])
                costsHorz[choose][to] = min(costsHorz[choose][to], diffHorz[from][to])
            }
        }
    }

    val answers = LongArray(n + 1) { Long.MAX_VALUE }

    for (b in 0 until (3).simplePowExact(n)) {
        var chooseVertBit = 0
        var chooseHorzBit = 0
        var bitOfChooseTypes = b
        var bitOfDigit2 = 1

        while (bitOfChooseTypes > 0) {
            when (bitOfChooseTypes % 3) {
                1 -> chooseVertBit = chooseVertBit or bitOfDigit2
                2 -> chooseHorzBit = chooseHorzBit or bitOfDigit2
            }
            bitOfDigit2 = bitOfDigit2 shl 1
            bitOfChooseTypes /= 3
        }

        val costSum = (0 until n)
            .map { min(costsVert[chooseVertBit][it], costsHorz[chooseHorzBit][it]).toLong() * pList[it] }.sum()

        val chooseCount = Integer.bitCount(chooseVertBit or chooseHorzBit)
        answers[chooseCount] = min(answers[chooseCount], costSum)
    }

    val answer = answers.joinToString(System.lineSeparator())

    println(answer)
}

private fun Int.simplePowExact(p: Int): Int {
    var x = 1
    repeat(p) { x = Math.multiplyExact(x, this) }
    return x
}