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: Accepted)

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


fun main(args: Array<String>) {
    Program().solve()
}

class Program {
    fun solve() {
        val sc = MyScanner()
        val n = sc.nextInt()
        val x = Array(n) { 0L }
        val y = Array(n) { 0L }
        val p = Array(n) { 0L }
        for (i in 0 until n) {
            x[i] = sc.nextLong()
            y[i] = sc.nextLong()
            p[i] = sc.nextLong()
        }

        val cost = Array(n + 1) { Array(n) { Long.MAX_VALUE } }
        var sum = 0L
        // 線路0本
        for (i in 0 until n) {
            cost[0][i] = min(abs(x[i]), abs(y[i])) * p[i]
            sum += cost[0][i]
        }

        val ans = Array(n + 1) { Long.MAX_VALUE }

        fun go(t: Int, k: Int) {
            ans[k] = min(ans[k], sum)
            if (t >= n) return
            // 縦
            var diff = 0L
            for (i in 0 until n) {
                cost[t + 1][i] = min(cost[t][i], abs(x[i] - x[t]) * p[i])
                diff += cost[t + 1][i] - cost[t][i]
            }

            sum += diff
            go(t + 1, k + 1)
            sum -= diff
            diff = 0

            // 横
            for (i in 0 until n) {
                cost[t + 1][i] = min(cost[t][i], abs(y[i] - y[t]) * p[i])
                diff += cost[t + 1][i] - cost[t][i]
            }

            sum += diff
            go(t + 1, k + 1)
            sum -= diff

            for (i in 0 until n) {
                cost[t + 1][i] = cost[t][i]
            }

            go(t + 1, k)
        }

        go(0, 0)

        println(ans.joinToString("\n"))
    }
}

class MyScanner {
    private var ptr = 0
    private var l = listOf<String>()

    fun next(): String {
        while (l.size <= ptr) {
            l = readLine()!!.split(' ')
            ptr = 0
        }
        return l[ptr++]
    }

    fun nextInt() = next().toInt()
    fun nextLong() = next().toLong()
    fun nextDouble() = next().toDouble()
    fun nextCharArray() = next().toCharArray()
    fun nextChar() = next()[0]

    fun list(): List<String> {
        ptr = l.size
        return readLine()!!.split(' ')
    }

    fun intList() = list().map { s -> s.toInt() }
    fun longList() = list().map { s -> s.toLong() }
    fun doubleList() = list().map { s -> s.toDouble() }

    fun array() = list().toTypedArray()
    fun intArray() = intList().toIntArray()
    fun longArray() = longList().toLongArray()
    fun doubleArray() = doubleList().toDoubleArray()
}