Problem: 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.

Examples

Input

3
1 2 300
3 3 600
1 4 800


Output

2900
900
0
0


Input

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400


Output

13800
1600
0
0
0
0


Input

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200


Output

26700
13900
3200
1200
0
0
0


Input

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


Output

2576709
1569381
868031
605676
366338
141903
0
0
0

Solution (in PYTHON3):

Solution:
# Fast IO (only use in integer input)

# TLE

import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

def main():

    N = int(input())
    Resident = []
    for i in range(N):
        x,y,p = map(int,input().split())
        Resident.append((x,y,p))

    minCost = [float("inf")] * (N+1)

    preCalRange = 1 << N

    horizontalLineDistance = []
    verticalLineDistance = []

    for i in range(preCalRange):
        railRoad = bin(i)[2:]
        railRoad = '0' * (N - len(railRoad)) + railRoad
        horizontalLineDistance.append([])
        verticalLineDistance.append([])
        for j in range(N):
            horizontalLineDistance[i].append(min(abs(Resident[j][0]),abs(Resident[j][1])))
            verticalLineDistance[i].append(min(abs(Resident[j][0]),abs(Resident[j][1])))
            for k in range(N):
                if railRoad[k] == '1' and abs(Resident[j][0] - Resident[k][0]) < horizontalLineDistance[i][j]:
                    horizontalLineDistance[i][j] = abs(Resident[j][0] - Resident[k][0])
                if railRoad[k] == '1' and abs(Resident[j][1] - Resident[k][1]) < verticalLineDistance[i][j]:
                    verticalLineDistance[i][j] = abs(Resident[j][1] - Resident[k][1])

    power2 = [1]
    for i in range(N):
        power2.append(power2[-1] * 2)
    loopCount = power2[N]
    for i in range(loopCount):
        numNewRailroad = bin(i)[2:].count('1')
        j = i
        while j >= 0:
            j &= i
            goodList1 = horizontalLineDistance[j]
            goodList2 = verticalLineDistance[i - j]
            cost = 0
            for k in range(N):
                cost += Resident[k][2] * min(goodList1[k], goodList2[k])
            if minCost[numNewRailroad] > cost:
                minCost[numNewRailroad] = cost
            j -= 1
    for elem in minCost:
        print(elem)

main()