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: Time Limit Exceeded)

#include"bits/stdc++.h"

using namespace std; 

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef long long LL;
typedef vector<LL> VL;
typedef vector<VL> VVL;
typedef pair<LL, LL> PLL;
typedef vector<PLL> VPLL;
typedef priority_queue<LL> PQ_DESC;
typedef priority_queue<LL, VL, greater<LL>> PQ_ASC;
typedef priority_queue<PII> PQ_DESC_PII;
typedef priority_queue<PII, vector<PII>, greater<PII>> PQ_ASC_PII;
typedef priority_queue<VL> PQ_DESC_VL;
typedef priority_queue<VL, vector<VL>, greater<VL>> PQ_ASC_VL;
typedef priority_queue<PLL> PQ_DESC_PLL;
typedef priority_queue<PLL, vector<PLL>, greater<PLL>> PQ_ASC_PLL;

#define ALL(c) (c).begin(),(c).end()
#define PB push_back
#define MP make_pair
#define SORT_ASC(c) sort(ALL(c))
//#define SORT_DESC(c) sort(ALL(c), greater<typeof(*((c).begin()))>())
#define SORT_DESC(c) sort((c).rbegin(),(c).rend())
#define REV(c) reverse((c).begin(), (c).end())
#define SIZE(a) int((a).size())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define ROF(i,a,b) for(int i=(b-1);i>=(a);--i)
#define REP(i,n) FOR(i,0,n)
#define PER(i,n) ROF(i,0,n)

const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int LARGE_INT = 1e9+100;
const int INF = 2e9+100;
const LL INF_LL = (LL)INF*(LL)INF;
const int MOD = 1e9+7;

//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

LL modpow(LL a, LL n) {
    LL res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}

int n;
VL result;
VVL mati;
vector<map<VL,LL>> memo;

LL calc(int depth, VL douro, set<LL> x, set<LL> y){
    LL d=0;
    REP(i,n){
        LL min_x;
        auto itx1 = x.lower_bound(mati[i][0]);
        auto itx2 = x.lower_bound(mati[i][0]);
        if(itx1 == x.end()){
            itx2--;
            min_x = abs((*itx2) - mati[i][0]); 
        }
        else if(itx1 == x.begin()){
            min_x = abs((*itx1) - mati[i][0]);
        }
        else{
            itx2--;
            min_x = min(abs((*itx1) - mati[i][0]), abs((*itx2) - mati[i][0]));
        }
        LL min_y;
        auto ity1 = y.lower_bound(mati[i][1]);
        auto ity2 = y.lower_bound(mati[i][1]);
        if(ity1 == y.end()){
            ity2--;
            min_y = abs((*ity2) - mati[i][1]); 
        }
        else if(ity1 == y.begin()){
            min_y = abs((*ity1) - mati[i][1]);
        }
        else{
            ity2--;
            min_y = min(abs((*ity1) - mati[i][1]), abs((*ity2) - mati[i][1]));
        }

        d += min(min_x,min_y) * mati[i][2];
    }

    return d;
}

void dfs(int depth, VL douro){
    LL d=0;
    REP(i,n-1){
        LL small = INF_LL;
        REP(j,n){
            if(douro[j]==0){
                continue;
            }
            else if(douro[j]==1){
                small = min(small,abs(mati[i][0]-mati[j][0]));
            }
            else if(douro[j]==2){
                small = min(small,abs(mati[i][1]-mati[j][1]));
            }
            else if(douro[j]==3){
                small = min(small,abs(mati[i][0]-mati[j][0]));
                small = min(small,abs(mati[i][1]-mati[j][1]));
            }
        }
        d += small * mati[i][2];
    }

    result[depth] = min(result[depth],d);
    memo[depth].insert({douro,d});


    if(depth==n){
        return;
    }

    REP(i,n){
        if(douro[i]==0){
            VL next_douro = douro;
            next_douro[i] = 1;
            if(memo[depth+1].find(next_douro) == memo[depth+1].end()){
                dfs(depth+1,next_douro);
                //continue;
            }
            next_douro[i] = 2;
            if(memo[depth+1].find(next_douro) == memo[depth+1].end()){
                dfs(depth+1,next_douro);
                //continue;
            }

            //LL dx = calc(depth+1,next_douro,next_x,y);
            //LL dy = calc(depth+1,next_douro,x,next_y);

        }
    }
    return;   
}

void Main()
{
    cin>>n;

    mati.resize(n);
    REP(i,n){
        LL x,y,p;cin>>x>>y>>p;
        mati[i] = {x,y,p};
    }
    mati.PB({0,0,0});
    n++;


    result.resize(n, INF_LL);
    memo.resize(n);
    VL douro(n);
    douro[n-1]=3;

    dfs(0,douro);

    REP(i,n){
        cout<<result[i]<<endl;
    }


	return;
}

int main()
{
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(15);
	Main();
    return 0;
}