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 java.io.BufferedReader
import kotlin.math.absoluteValue
import kotlin.math.min

fun main() {
    val br = System.`in`.bufferedReader()
    val sb = StringBuilder()
    val n = br.readInt()
    val areas = Array<Triple<Int, Int, Long>>(n) { Triple(br.readInt(), br.readInt(), br.readInt().toLong()) }
    val lows = LongArray(n + 1) { Long.MAX_VALUE }
    fun solve(state: Int, numAdded: Int, nextToAdd: Int) {
        if (nextToAdd == n) {
            var cost = 0L
            for (city in 0 until n) {
                var cityCost = min(areas[city].first.absoluteValue, areas[city].second.absoluteValue)
                for (road in 0 until n) {
                    when ((state shr (road * 2)) and 3) {
                        2 -> {
                            cityCost = cityCost.coerceAtMost((areas[road].first - areas[city].first).absoluteValue)
                        }
                        1 -> {
                            cityCost = cityCost.coerceAtMost((areas[road].second - areas[city].second).absoluteValue)
                        }
                    }
                }
                cost += cityCost * areas[city].third
            }
            lows[numAdded] = min(lows[numAdded], cost)
        } else {
            solve(state, numAdded, nextToAdd + 1)
            for (x in 1..2) {
                solve(state or (x shl (nextToAdd * 2)), numAdded + 1, nextToAdd + 1)
            }
        }
    }
    solve(0, 0, 0)
    for (k in 0..n) {
        sb.append(lows[k]).append('\n')
    }
    print(sb)
}


private const val SPACE_INT = ' '.toInt()
private const val ZERO_INT = '0'.toInt()
private const val NL_INT = '\n'.toInt()

private fun BufferedReader.readInt(): Int {
    var ret = read()
    while (ret <= SPACE_INT) {
        ret = read()
    }
    val neg = ret == '-'.toInt()
    if (neg) {
        ret = read()
    }
    ret -= ZERO_INT
    var read = read()
    while (read >= ZERO_INT) {
        ret *= 10
        ret += read - ZERO_INT
        read = read()
    }

    while (read <= SPACE_INT && read != -1 && read != NL_INT) {
        mark(1)
        read = read()
    }
    if (read > SPACE_INT) {
        reset()
    }
    return if (neg) -ret else ret
}