Problem:
Score: 500 points

Problem Statement

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.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

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

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//コンパイラ最適化用
#pragma GCC optimize("Ofast")
//インクルード(アルファベット順)
#include<algorithm>//sort,二分探索,など
#include<bitset>//固定長bit集合
#include<chrono>//実行時間計測
#include<cmath>//pow,logなど
#include<complex>//複素数
#include<deque>//両端アクセスのキュー
#include<functional>//sortのgreater
#include<iomanip>//setprecision(浮動小数点の出力の誤差)
#include<iostream>//入出力
#include<iterator>//集合演算(積集合,和集合,差集合など)
#include<map>//map(辞書)
#include<numeric>//iota(整数列の生成),gcdとlcm,accumulate
#include<queue>//キュー
#include<set>//集合
#include<stack>//スタック
#include<string>//文字列
#include<unordered_map>//順序保持しないmap
#include<unordered_set>//順序保持しないset
#include<utility>//pair
#include<vector>//可変長配列
#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, int> pii;
typedef pair<ll, ld> pld;
typedef pair<pii, int> ppiii;
typedef pair<pii, ll> ppiill;
typedef pair<pll, ll> ppllll;
typedef pair<pli, int> pplii;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<int>> vvint;
typedef vector<vector<pll>> vvpll;

//マクロ
//forループ
//引数は、(ループ内変数,動く範囲)か(ループ内変数,始めの数,終わりの数)、のどちらか
//Dがついてないものはループ変数は1ずつインクリメントされ、Dがついてるものはループ変数は1ずつデクリメントされる
#define REP(i,n) for(ll i=0;i<ll(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=ll(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=ll(b);i--)
//xにはvectorなどのコンテナ
#define ALL(x) x.begin(),x.end() 
#define SIZE(x) ll(x.size()) 
//定数
#define INF 1000000000000 //10^12:∞
#define MOD 1000000007 //10^9+7:合同式の法
#define MAXR 100000 //10^5:配列の最大のrange
//略記
#define PB push_back //挿入
#define MP make_pair //pairのコンストラクタ
#define F first //pairの一つ目の要素
#define S second //pairの二つ目の要素
#define cin(x) cin >> x

vector<long long> solve(int N, vector<int> X, vector<int> Y, vector<int> P){
    const ll inf = 1LL <<62;
    int mul = 1;
    REP(i,N){mul*= 3;}
    vvint xsel(1<<N, vint(N));
    vvint ysel(1<<N, vint(N));
    for(int i =0; i <1 <<N; ++i){
        REP(j,N){
            xsel[i][j] = abs(X[j]);
            ysel[i][j] = abs(Y[j]);
            REP(k,N){
                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])); 
                }
            }
        }
    }
    vll ans(N+1, inf);
    for(int i =0; i < 1 << N; ++i){
        int cnt =0;
        REP(j,N){
            if((i >> j) & 1) ++cnt;
        }
        for(int j =i; j >= 0; --j){
            j &= i;
            ll cost = 0;
            REP(k,N){
                if(!((i >> k) & 1)){
                    cost += 1LL * min({xsel[j][k], ysel[i -j][k]}) *P[k];
                }
            }
            ans[cnt] = min(ans[cnt],cost);
        }
    }
    return ans;
}

signed main(){
    //入力の高速化用のコード
    //ios::sync_with_stdio(false);
    //cin.tie(nullptr);
    int N=0;
    cin(N);
    vint X(N), Y(N), P(N); 
    for(int i=0; i< N; i++){
        cin >> X[i] >> Y[i] >> P[i];
    }

    vll ans = solve(N, X ,Y , P);
    REP(i,N+1){
        cout << ans[i] << endl;
    }

  return 0;
}