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:
import bisect
import copy
import heapq
import sys
import itertools
import math
from functools import lru_cache
from collections import deque
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
mod = 10 ** 9 + 7 

def read_values(): return map(int, input().split())
def read_index(): return map(lambda x: int(x) - 1, input().split())
def read_list(): return list(read_values())
def read_lists(N): return [read_list() for n in range(N)]


def main():
    N = int(input())
    P = read_lists(N)
    # N = 15
    # P = [(i ** 2 + 1, i ** 2 + 1, i ** 2 + 1) for i in range(N)]
    
    C = [1 << 62] * (N + 1)
    # CX = [None for _ in range(2 ** N)]
    # CY = [None for _ in range(2 ** N)]
    CX = [1 << 62] * (N * 2 ** N)
    CY = [1 << 62] * (N * 2 ** N)

    LX = [0] * (2 ** N)
    LY = [0] * (2 ** N)
    for K in range(2 ** N):
        X = {0}
        Y = {0}
        for i in range(N):
            if (K >> i) & 1:
                X.add(P[i][0])
                Y.add(P[i][1])
         
        # CXX = [0] * N
        # CYY = [0] * N
        for j, (xp, yp, cp) in enumerate(P):
            for x in X:
                # CXX[j] = min(abs(xp - x) * cp, CXX[j])
                CX[K * N + j] = min(abs(xp - x) * cp, CX[K * N + j])
            for y in Y:
                CY[K * N + j] = min(abs(yp - y) * cp, CY[K * N + j])
                # CYY[j] = min(abs(yp - y) * cp, CYY[j])
            
        LX[K] = len(X)
        LY[K] = len(Y)
        # CX[K] = tuple(CXX)
        # CY[K] = tuple(CYY)

    for T in range(2 ** N):
        p = format(T, "160b").count("1")
        if C[p] == 0:
            continue

        r = 1 << 62
        tx = T
        while True:
            ty = T - tx
            c = 0
            for n in range(N):
                c += min(CX[tx * N + n], CY[ty * N + n])
            r = min(r, c)
            if tx == 0:
                break
            tx = (tx - 1) & T
        C[p] = min(C[p], r)
    print("\n".join(map(str, C)))


if __name__ == "__main__":
    main()