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:
def main():
    n = int(input())
    xyp = [list(map(int, input().split())) for _ in [0]*n]
    d = []

    def calc(seq):
        m = len(seq)
        dp = [[0] for _ in [0]*1]  # 使った回数、前のやつ
        for i in range(m):
            dp2 = [[10**15]*(i+2) for _ in [0]*(i+2)]
            for j in range(i+1):
                for k in range(i+1):
                    dp2[j][k] = min(dp2[j][k], dp[j][k])
                    temp = dp[j][k]
                    if k == 0:
                        for l in range(k, i+1):
                            temp += min(abs(seq[l][0]), seq[i]
                                        [0]-seq[l][0])*seq[l][1]
                    else:
                        for l in range(k, i+1):
                            temp += min(abs(seq[l][0]), seq[l]
                                        [0]-seq[k-1][0], seq[i][0]-seq[l][0])*seq[l][1]
                    dp2[j+1][i+1] = min(dp2[j+1][i+1], temp)
            dp = dp2
        for i in range(m+1):
            for l in range(m):
                dp[i][0] += abs(seq[l][0])*seq[l][1]
            for k in range(1, m+1):
                for l in range(k, m):
                    dp[i][k] += min(abs(seq[l][0]), seq[l]
                                    [0]-seq[k-1][0])*seq[l][1]
        for i in range(m+1):
            dp[i] = min(dp[i])
        dp += [0]*(n-m)
        return dp

    xyp.sort()
    for i in range(2**n):
        x_list, y_list = [], []
        for j in range(n):
            x, y, p = xyp[j]
            if i & 1:
                x_list.append([x, p])
            else:
                y_list.append([y, p])
            i >>= 1
        y_list.sort()
        d.append([calc(x_list), calc(y_list)])

    ans = [10**15]*(n+1)
    for x, y in d:
        for i in range(n+1):
            for j in range(n-i+1):
                ans[i+j] = min(ans[i+j], x[i]+y[j])
    for i in ans:
        print(i)


main()