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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> x(n);
    vector<int> y(n);
    vector<long long> p(n);
    for(int i=0; i<n; i++) cin >> x[i] >> y[i] >> p[i];

    //初期化
    int ax[33000][15];
    int ay[33000][15];
    long long c[20];
    for(int i=0; i<=n; i++) c[i] = 1000000000000000000;

     for(int bit = 0; bit<(1<<n); bit++){
        for(int i=0; i<n; i++){
            ax[bit][i] = abs(x[i]);
            for(int j=0; j<n; j++){
                if((bit >> j) & 1) ax[bit][i] = min(ax[bit][i], abs(x[i]-x[j]));
            }
        }
     }

    for(int bit = 0; bit<(1<<n); bit++){
        for(int i=0; i<n; i++){
            ay[bit][i] = abs(y[i]);
            for(int j=0; j<n; j++){
                if((bit >> j) & 1) ay[bit][i] = min(ay[bit][i], abs(y[i]-y[j]));
            }
        }
     }
     
     //3進数bit全探索
     for (int bit = 0; bit<pow(3,n); bit++){
         int tmp = bit;
         int bitx=0;
         int bity=0;
         int k=0;
         for (int pos = 0; pos<n; pos++){
             if(tmp%3==1){
                 bitx += (1<<pos);
                 k++;
                 }
            if(tmp%3==2){
                 bity += (1<<pos);
                 k++;
                 }
             tmp /= 3;
        }
        long long t=0;
        for(int i=0; i<n; i++){
            t += min(ax[bitx][i],ay[bity][i])*p[i];
        }
        c[k] = min(c[k], t);
     }

    for(int k=0; k<=n; k++) cout << c[k] << endl;
}