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:
N = int(input())
X = []
Y = []
P = []
for _ in range(N):
    x, y, p = map(int, input().split())
    X.append(x)
    Y.append(y)
    P.append(p)

DX = [[0] * (1 << N) for _ in range(N)]
DY = [[0] * (1 << N) for _ in range(N)]

for i, x in enumerate(X):
    for k in range(1 << N):
        d = abs(x)
        for j, xx in enumerate(X):
            if k & (1 << j):
                d = min(d, abs(xx - x))
        DX[i][k] = d * P[i]
for i, y in enumerate(Y):
    for k in range(1 << N):
        d = abs(y)
        for j, yy in enumerate(Y):
            if k & (1 << j):
                d = min(d, abs(yy - y))
        DY[i][k] = d * P[i]

ANS = [1 << 100] * (N + 1)
PC = [0] * (1 << N)
for i in range(1, 1 << N):
    PC[i] = PC[i^1] + 1 if i % 2 else PC[i//2]

RA = []
for i in range(1 << N):
    t = []
    for j in range(N):
        if i & (1 << j):
            t.append(j)
    RA.append(t)

mm = (1 << N) - 1
for i in range(1 << N):
    m = i ^ mm
    j = m
    while 1:
        ans = 0
        for k in RA[i^j^mm]:
            ans += min(DX[k][i], DY[k][j])
        
        pc = PC[i] + PC[j]
        ANS[pc] = min(ANS[pc], ans)
        
        if not j:
            break
        j = m & (j - 1)

print(*ANS, sep = "\n")