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;
using ll = long long;

vector<ll> solve(int N, vector<int> X, vector<int> Y, vector<int> P){
    const ll inf = 1LL << 62;
    vector<vector<int>> xsel(1 << N, vector<int>(N));
    vector<vector<int>> ysel(1 << N, vector<int>(N));
    for(int i = 0; i < 1 << N; i++){
        for(int j = 0; j < N; j++){
            xsel[i][j] = abs(X[j]);
            ysel[i][j] = abs(Y[j]);
            for(int k = 0; k < N; k++){
                if((i >> k) & 1){
                    xsel[i][j] = min(xsel[i][j], abs(X[j] - X[k]));
                    ysel[i][j] = min(ysel[i][j], abs(Y[j] - Y[k]));
                }
            }
        }
    }
    vector<ll> ans(N+1, inf);
    for(int i = 0; i < 1 << N; i++){
        int cnt = 0;
        for(int j = 0; j < N; j++){
            if((i >> j) & 1) cnt++;
        }
        for(int j = i; j >= 0; j--){
            j &= i;
            ll cost = 0;
            for(int k = 0; k < N; k++){
                if(!((i >> k) & 1)){
                    cost += 1LL * min(xsel[j][k], ysel[i-j][k]) * P[k];
                }
            }
            ans[cnt] = min(ans[cnt], cost);
        }
    }
    return ans;
}

int main(){
    int N;
    cin >> N;
    vector<int> X(N), Y(N), P(N);
    for(int i = 0; i < N; i++){
        cin >> X[i] >> Y[i] >> P[i];
    }
    vector<ll> ans = solve(N, X, Y, P);
    for(int i = 0; i <= N; i++){
        cout << ans[i] << endl;
    }
}