#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
typedef int64_t Int;
typedef pair<Int, Int> P;

vector<int> pow3 = {1};

int main(){
    int N; cin >> N;
    vector<Int> X(N), Y(N), P(N);
    rep(i,N) cin >> X[i] >> Y[i] >> P[i];

    rep(i,N) pow3.push_back(pow3.back() * 3);
    
    // 前処理：縦の路線・横の路線しか建てないときの各集落からの歩行距離
    vector<vector<Int>> V_only(1<<N, vector<Int>(N));
    vector<vector<Int>> H_only(1<<N, vector<Int>(N));
    rep(bit, 1<<N){
        rep(i,N){ // 集落 i からの歩行距離を計算
            Int dist_V = P[i] * abs(X[i]);
            Int dist_H = P[i] * abs(Y[i]);

            rep(j,N){
                if(bit>>j & 1){
                    dist_V = min(dist_V, P[i] * abs(X[i] - X[j]));
                    dist_H = min(dist_H, P[i] * abs(Y[i] - Y[j]));
                } 
            }

            V_only[bit][i] = dist_V;
            H_only[bit][i] = dist_H;
        }
    }

    // 全ての 3^N 通りの道路の建て方を試す．
    vector<Int> ans(N+1, 1e18);
    rep(trit, pow3[N]){
        int K = 0;

        int bitV = 0, bitH = 0;
        rep(i,N){
            if(trit / pow3[i] % 3 == 1){ 
                bitV += 1<<i; // X[i] 上に道を建てる
                K++;
            } else if (trit / pow3[i] % 3 == 2){
                bitH += 1<<i; // Y[i] 上に道を建てる
                K++;
            }
        }

        Int dist = 0;
        rep(i,N) dist += min(V_only[bitV][i], H_only[bitH][i]);
        
        ans[K] = min(ans[K], dist);
    }

    for(Int x : ans) cout << x << endl;
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
typedef int64_t Int;
typedef pair<Int, Int> P;

vector<int> pow3 = {1};

int main(){
    int N; cin >> N;
    vector<Int> X(N), Y(N), P(N);
    rep(i,N) cin >> X[i] >> Y[i] >> P[i];

    rep(i,N) pow3.push_back(pow3.back() * 3);
    
    // 前処理：縦の路線・横の路線しか建てないときの各集落からの歩行距離
    vector<vector<Int>> V_only(1<<N, vector<Int>(N));
    vector<vector<Int>> H_only(1<<N, vector<Int>(N));
    rep(bit, 1<<N){
        rep(i,N){ // 集落 i からの歩行距離を計算
            Int dist_V = P[i] * abs(X[i]);
            Int dist_H = P[i] * abs(Y[i]);

            rep(j,N){
                if(bit>>j & 1){
                    dist_V = min(dist_V, P[i] * abs(X[i] - X[j]));
                    dist_H = min(dist_H, P[i] * abs(Y[i] - Y[j]));
                } 
            }

            V_only[bit][i] = dist_V;
            H_only[bit][i] = dist_H;
        }
    }

    // 全ての 3^N 通りの道路の建て方を試す．
    vector<Int> ans(N+1, 1e18);
    rep(trit, pow3[N]){
        int K = 0;

        int bitV = 0, bitH = 0;
        rep(i,N){
            if(trit / pow3[i] % 3 == 1){ 
                bitV += 1<<i; // X[i] 上に道を建てる
                K++;
            } else if (trit / pow3[i] % 3 == 2){
                bitH += 1<<i; // Y[i] 上に道を建てる
                K++;
            }
        }

        Int dist = 0;
        rep(i,N) dist += min(V_only[bitV][i], H_only[bitH][i]);
        
        ans[K] = min(ans[K], dist);
    }

    for(Int x : ans) cout << x << endl;
}

```