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 i in range(n)]

    def caldp(a):
        n = len(a)
        a.sort()
        dp = [[float("INF")]*(n+1) for i in range(n+1)]
        dp[0][0] = 0
        for l in range(n):
            for r in range(l+1,n+1):
                now = float("INF")
                for k in range(l,r):
                    count = 0
                    for i in range(l,r):
                        count += abs(a[i][0]-a[k][0])*a[i][1]

                    now = min(now,count)
                for j in range(n):
                    dp[r][j+1] = min(dp[r][j+1],dp[l][j]+now)

                now = 0
                for i in range(l,r):
                    now += abs(a[i][0])*a[i][1]
                for j in range(n+1):
                    dp[r][j] = min(dp[r][j],dp[l][j]+now)

        res = [float("INF")]*(n+1)
        for i in range(n+1):
            res[i] = dp[-1][i]
        return res

    ans = [float("INF")]*(n+1)

    for s in range(1<<n):
        lx = []
        ly = []
        for i in range(n):
            if s >> i & 1:
                lx.append((xyp[i][0],xyp[i][2]))
            else:
                ly.append((xyp[i][1],xyp[i][2]))
        lx = caldp(lx)
        ly = caldp(ly)
        for i in range(len(lx)):
            for j in range(len(ly)):
                ans[i+j] = min(ans[i+j],lx[i]+ly[j])

    for i in ans:
        print(i)

if __name__ == "__main__":
    main()