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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

def main():
    from bisect import bisect_right as br
    from itertools import combinations
    n = int(input())
    xyp = [list(map(int, input().split())) for _ in [0]*n]
    d = []

    def median(list1):
        w_list = [0]
        for x, w in list1:
            w_list.append(w_list[-1]+w)
        w = w_list[-1]/2
        b = br(w_list, w)
        m = list1[b-1][0]
        ans = 0
        for x, w in list1:
            ans += abs(x-m)*w
        return ans

    cb = [[list(combinations([k+1 for k in range(j)], i))
           for j in range(n+1)] for i in range(n+1)]
    for i in range(n+1):
        for j in range(n+1):
            for k in range(len(cb[i][j])):
                cb[i][j][k] = [0]+list(cb[i][j][k])+[j+1]

    def calc(list_temp):
        list1 = [i for i in list_temp]
        ans = [10**15]*(n+1)
        m = len(list1)
        zero_sum = 0
        # 何個0で使うか
        for zero in range(m+1):
            calc2(list1, ans, zero_sum)
            if list1:
                zero_sum += abs(list1[-1][0])*list1[-1][1]
                list1.pop()
        ans[0] = zero_sum
        return ans

    def calc2(list2, ans, zero_sum):
        list1 = sorted([i for i in list2])
        m = len(list1)
        # i=1のときは別途
        if m <= 1:
            ans[1] = min(ans[1], zero_sum)
        else:
            ans[1] = min(ans[1], zero_sum+median(list1))
        # i=2以降
        for i in range(2, n+1):
            if m <= i:
                ans[i] = min(ans[i], zero_sum)
                continue
            ans3 = 10**15
            for c in cb[i-1][m-1]:
                ans2 = zero_sum
                end = 0
                for j in c[1:i+1]:
                    start = end
                    end = j
                    ans2 += median(list1[start:end])
                ans3 = min(ans3, ans2)
            ans[i] = min(ans[i], ans2)

    xyp.sort()
    for i in range(2**n):
        x_list2, y_list2 = [], []
        for j in range(n):
            x, y, p = xyp[j]
            if i & 1:
                x_list2.append([x, p])
            else:
                y_list2.append([y, p])
            i >>= 1
        y_list2.sort()
        x_list = x_list2[:1]
        y_list = y_list2[:1]
        for j in range(1, len(x_list2)):
            if x_list2[j][0] == x_list[-1][0]:
                x_list[-1][1] += x_list2[j][1]
            else:
                x_list.append(x_list2[j])
        for j in range(1, len(y_list2)):
            if y_list2[j][0] == y_list[-1][0]:
                y_list[-1][1] += y_list2[j][1]
            else:
                y_list.append(y_list2[j])
        x_list.sort(key=lambda x: (-abs(x[0]), -x[1]))
        y_list.sort(key=lambda x: (-abs(x[0]), -x[1]))
        d.append([calc(x_list), calc(y_list)])

    ans = [10**15]*(n+1)
    for x, y in d:
        for i in range(n+1):
            for j in range(n-i+1):
                ans[i+j] = min(ans[i+j], x[i]+y[j])
    for i in ans:
        print(i)


main()