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: Wrong Answer)

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

fun main() {
    val n = nextInt()
    val X = mutableListOf<Int>()
    val Y = mutableListOf<Int>()
    val P = mutableListOf<Int>()
    repeat(n) {
        val (x, y, p) = nextIntList()
        X.add(x)
        Y.add(y)
        P.add(p)
    }
    val xsel = Array(1 shl n) { Array(n) { 0 } }
    val ysel = Array(1 shl n) { Array(n) { 0 } }
    (0 until (1 shl n)).forEach { i ->
        (0 until n).forEach { j ->
            xsel[i][j] += abs(X[j])
            ysel[i][j] += abs(Y[j])
            (0 until n).forEach { k ->
                if ((i shr k) and 1 == 1) {
                    xsel[i][j] = min(xsel[i][j], abs(X[j] - X[k]))
                    ysel[i][j] = min(ysel[i][j], abs(Y[j] - Y[k]))
                }
            }
        }
    }
    (0 until (1 shl n)).forEach { i ->
        (0 until n).forEach { j ->
            xsel[i][j] *= P[j]
            ysel[i][j] *= P[j]
        }
    }

    val ans = sizedArray(n + 1, Long.MAX_VALUE)
    (0 until (1 shl n)).forEach { i ->
        val count = (0 until n).count {
            i shr it and 1 == 1
        }
        var j = i
        while (j >= 0) {
            j = j and i
            var cost = 0L
            (0 until n).forEach { k ->
                if ((i shr k) and 1 != 1) {
                    cost += min(xsel[j][k], ysel[i - j][k])
                }
            }
            ans[count] = min(ans[count], cost)
            j--
        }
//        ans.forEach(::println)
//        return
    }
    ans.forEach(::println)
}

// # Utils
// ## Input
fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun nextList() = next().split(" ")
fun nextIntList() = next().split(" ").map(String::toInt)
fun nextLongList() = next().split(" ").map(String::toLong)
fun nextDoubleList() = next().split(" ").map(String::toDouble)

// ## Array
inline fun <reified T> sizedArray(size: Int, default: T) = Array(size) { default }

// ## mod^-1
//    [NOTE] CANNOT use this for 0, mod, 2mod, 3mod ...(return wrong value)
fun modinv(num: Long, mod: Long): Long {
    var a = num
    var b = mod
    var u = 1L
    var v = 0L
    while (b > 0) {
        val t = a.div(b)
        a -= t * b
        a = b.also { b = a } // swap a and b
        u -= t * v
        u = v.also { v = u } // swap u and v
    }
    u %= mod
    if (u < 0) u += mod
    return u
}