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: Time Limit Exceeded)

def score(CITIES, lines, lower_bound, city_lines, city_scores):
    s = 0
    for cityid in range(len(CITIES)):
        x = CITIES[cityid][0]
        y = CITIES[cityid][1]
        p = CITIES[cityid][2]

        for ctl in city_lines[cityid]:
            if ctl in lines:
                s += city_scores[cityid][ctl]
                break
        if s >= lower_bound:
            return lower_bound
    return s

def recursive_score(K, CITIES, candidate_lines, selected, min_select, lower_bound, city_lines, city_scores):
    if K == len(CITIES):
        return 0

    if len(selected) == K:
        lines = [(0, "H"), (0, "V")]
        for s in selected:
            lines.append(candidate_lines[s])
        return score(CITIES, lines, lower_bound, city_lines, city_scores)

    min_score = lower_bound
    i = min_select
    while i < len(candidate_lines):
        if i % 2 == 0:
            next_min_select = i + 2 # 同じ都市に縦線と横線を両方置くのは無駄
        else:
            next_min_select = i + 1
        sel = selected + [i]
        s = recursive_score(K, CITIES, candidate_lines, sel, next_min_select, min_score, city_lines, city_scores)
        if s == 0:
            return 0
        min_score = min(min_score, s)
        i += 1

    return min_score

def main():
    N = int(input())
    CITIES = []

    for i in range(N):
        # x=c[0]
        # y=c[1]
        # p=c[2]
        c = [int(_x) for _x in input().split()]
        CITIES.append(c)

    candidate_lines_org = []
    for i in range(N):
        candidate_lines_org.append((CITIES[i][1], "H"))  # y座標が同じで横線
        candidate_lines_org.append((CITIES[i][0], "V"))  # x座標が同じで縦線
    candidate_lines = candidate_lines_org

    # 2つ以上の都市に乗る線は優先度を高くしつつ、一回のみカウント
    # duplicate_lines = []
    # single_lines = []
    # for i in range(len(candidate_lines_org)):
    #     dup = False
    #     for j in range(len(candidate_lines_org)):
    #         if j <= i:
    #             continue
    #         if candidate_lines_org[i] == candidate_lines_org[j]:
    #             dup = True
    #             duplicate_lines.append(candidate_lines_org[i])
    #             break
    #     if not dup:
    #         single_lines.append(candidate_lines_org[i])
    # candidate_lines = duplicate_lines + single_lines

    # 線-都市ペアでのscoreを事前計算
    score_target_lines = candidate_lines + [(0, "H"), (0, "V")]
    city_lines = {}
    city_scores = {}
    for i in range(N):
        cityid_score = []

        x = CITIES[i][0]
        y = CITIES[i][1]
        p = CITIES[i][2]
        for l in score_target_lines:
            if l[1] == "H":
                dist = abs(y-l[0])
            else:
                dist = abs(x-l[0])
            cityid_score.append((dist*p, l))

        city_lines[i] = []
        for score, line in sorted(cityid_score, key=lambda x:x[0]):
            city_lines[i].append(line)
            if i not in city_scores:
                city_scores[i] = {}
            city_scores[i][line] = score

    lower_bound = 1000000000000
    for K in range(N+1):
        if lower_bound == 0:
            S = 0
        else:
            S = recursive_score(K, CITIES, candidate_lines, [], 0, lower_bound, city_lines, city_scores)
        print(S)
        lower_bound = S


main()