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() {
    data class Town(val x: Long, val y: Long, val population: Long)

    val n = readLine()!!.toInt()
    val towns = List(n) {
        val (x, y, p) = readLine()!!.split(" ").map { it.toLong() }
        Town(x, y, p)
    }

    // 何もしてないのときのコスト
    val baseCosts = towns.map { (x, y, p) -> min(abs(x), abs(y)) * p }.toLongArray()
    // [0:横,1:縦][i番目基準で道を引いたとき][j番目の街の移動コストがいくつになるか]
    val memo = Array(2) { Array(n) { LongArray(n) } }

    for (i in 0..1) {
        for ((fromIndex, fromTown) in towns.withIndex()) {
            for ((toIndex, toTown) in towns.withIndex()) {
                val (fromX, fromY, _) = fromTown
                val (toX, toY, toPopulation) = toTown
                if (i == 0) {
                    memo[0][fromIndex][toIndex] = minOf(abs(toX), abs(toY), abs(fromY - toY)) * toPopulation
                } else if (i == 1) {
                    memo[1][fromIndex][toIndex] = minOf(abs(toX), abs(toY), abs(fromX - toX)) * toPopulation
                }
            }
        }
    }

    val answer = StringBuilder()
    answer.appendln(baseCosts.sum())

    for (chooseCount in 1 until n) {
        var minCost = Long.MAX_VALUE
        searchAllPatterns(arrayOf(0, 1, 2), n) { isChooses ->
            if (isChooses.count { it in 0..1 } != chooseCount) return@searchAllPatterns
            val costs = baseCosts.copyOf()

            for ((from, chooseTypes) in isChooses.withIndex()) {
                if (chooseTypes == 2) continue
                for (to in 0 until n) costs[to] = min(costs[to], memo[chooseTypes][from][to])
            }

            minCost = min(minCost, costs.sum())
        }
        answer.appendln(minCost)
    }

    answer.appendln(0)

    print(answer)
}

@Suppress("SameParameterValue")
private inline fun <reified T> searchAllPatterns(patterns: Array<T>, length: Int, action: (Array<T>) -> Unit) {
    val patternSize = patterns.size
    val pow = (1..length).fold(1L) { acc: Long, _: Int -> acc * patternSize }
    val allPatterns = if (pow in 1..Int.MAX_VALUE) pow.toInt() else error("$pow is too large!")
    for (i in 0 until allPatterns) {
        val array = Array(length) { patterns[0] }
        var div = 1
        for (j in 0 until length) {
            array[length - j - 1] = patterns[(i / div) % patternSize]
            div *= patternSize
        }
        action(array)
    }
}

// region kokokara template dayo (^o^)
@Suppress("unused", "ClassName", "SpellCheckingInspection", "ConvertToStringTemplate", "MemberVisibilityCanBePrivate")
private class IO_msolutions2020e(private val separator: String = System.lineSeparator()) {
    private val input = System.`in`
    private val buffer = ByteArray(1024)
    private var pointer = 0
    private var bufferLength = 0
    private val sb = StringBuilder()
    private fun Byte.isPrintable() = this in 33..126
    private fun Byte.isNumeric() = this in '0'.toByte()..'9'.toByte()
    private fun Byte.toNumVal() = if (this.isNumeric()) this - '0'.toByte() else error("$this is not numeric")

    private fun hasNextByte(): Boolean {
        return if (pointer < bufferLength) true else {
            pointer = 0
            bufferLength = input.read(buffer)
            bufferLength > 0
        }
    }

    private fun readByte(): Byte = if (hasNextByte()) buffer[pointer++] else -1
    private fun skipUnprintable() = run { while (hasNextByte() && !buffer[pointer].isPrintable()) pointer++ }
    private fun hasNext(): Boolean = run { skipUnprintable() }.run { hasNextByte() }
    private fun hasNextOrError() = if (!hasNext()) error("has no next element.") else Unit

    fun readString(): String {
        hasNextOrError()
        val sb = StringBuilder()
        var b = readByte()
        while (b.isPrintable()) {
            sb.appendCodePoint(b.toInt())
            b = readByte()
        }
        return sb.toString()
    }

    fun readLong(): Long {
        hasNextOrError()
        var n = 0L
        var negative = false
        var b = readByte()
        if (b == '-'.toByte()) {
            negative = true
            b = readByte()
        }
        if (!b.isNumeric()) error("$b is not numeric.")
        while (true) {
            when {
                b.isNumeric() -> n = n * 10 + b.toNumVal()
                b.toInt() == -1 || !b.isPrintable() -> return if (negative) -n else n
                else -> error("failed to parse. [n=$n, b=$b]")
            }
            b = readByte()
        }
    }

    fun readInt() = readLong()
        .let { if (it in Int.MIN_VALUE..Int.MAX_VALUE) it.toInt() else error("$it is not in range of Int.") }

    fun readDouble(): Double {
        var n = 0.0
        var div = 1.0
        var negative = false
        var b = readByte()
        if (b == '-'.toByte()) {
            negative = true
            b = readByte()
        }
        do n = n * 10 + b.toNumVal()
        while (run { b = readByte() }.run { b.isNumeric() })
        if (b == '.'.toByte()) {
            while (run { b = readByte() }.run { b.isNumeric() })
                n += b.toNumVal() / (run { div *= 10 }.run { div })
        }
        return if (negative) -n else n
    }

    fun readStringList(size: Int) = readList(size) { readString() }
    fun readIntList(size: Int) = readList(size) { readInt() }
    fun readLongList(size: Int) = readList(size) { readLong() }
    fun readDoubleList(size: Int) = readList(size) { readDouble() }
    fun readIntArray(size: Int) = IntArray(size) { readInt() }
    fun readLongArray(size: Int) = LongArray(size) { readLong() }
    fun readDoubleArray(size: Int) = DoubleArray(size) { readDouble() }
    inline fun <reified T> readArray(size: Int, init: () -> T) = Array(size) { init() }
    inline fun <T> readList(size: Int, init: () -> T) = List(size) { init() }

    fun println(): Unit = run { sb.append(separator) }
    fun print(any: Any): Unit = run { sb.append(any) }
    fun println(any: Any): Unit = run { sb.append(any.toString() + separator) }

    fun exec(code: IO_msolutions2020e.() -> Unit) = run {
        Thread(null, { code() }, "(^o^)", 128 * 1024 * 1024).apply { start() }.join()
    }.run { kotlin.io.print(sb) }

    fun readLine(): Nothing = error("readLine is disabled.")
}
// endregion kokomade template dayo (^o^)