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: Accepted)

import itertools

N = int(input())

data = []
min_value_0 = 0
for _ in range(N):
	X, Y, P = town = tuple(map(int, input().split()))
	data.append(town)
	min_value_0 += min(abs(X), abs(Y)) * P

print(min_value_0)

B = 2 ** N
table_h = []
table_v = []

for b in range(B):
	towns = []
	i = 0
	x = b
	while x > 0:
		if x % 2 == 1:
			towns.append(i)
		x //= 2
		i += 1
	dists_h = []
	dists_v = []
	for i in range(N):
		X0, Y0, P0 = data[i]
		d_h = abs(X0)
		d_v = abs(Y0)
		for j in towns:
			X1, Y1, P1 = data[j]
			d_h = min(d_h, abs(X1 - X0))
			d_v = min(d_v, abs(Y1 - Y0))
		dists_h.append(d_h * P0)
		dists_v.append(d_v * P0)
	table_h.append(dists_h)
	table_v.append(dists_v)
#	print(f"{{:0{N}b}}".format(b), dist_h, dist_v)

table = [min_value_0] * (N + 1)

import sys

stack = [0] * N
i = 0
c = 0
b0 = 0
b1 = 0
b = 1
while True:
	if i == N:
		value = 0
		t1 = table_h[b0]
		t2 = table_v[b1]
		for j in range(N):
			value += min(t1[j], t2[j])
		if table[c] > value:
			table[c] = value
		i -= 1
		b //= 2
	else:
		s = stack[i]
		if s == 3:
			if i == 0:
				break
			else:
				stack[i] = 0
				i -= 1
				b //= 2
		else:
			if s == 0:
				b0 += b
				c += 1
			elif s == 1:
				b0 -= b
				b1 += b
			else:
				b1 -= b
				c -= 1
			stack[i] = s + 1
			i += 1
			b *= 2

for i in range(1, N):
	print(table[i])

print(0)