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)

#ifdef LOCAL
//#define _GLIBCXX_DEBUG
#endif
//#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<int, int> Pi;
typedef vector<ll> Vec;
typedef vector<int> Vi;
typedef vector<string> Vs;
typedef vector<char> Vc;
typedef vector<P> VP;
typedef vector<vector<ll>> VV;
typedef vector<vector<int>> VVi;
typedef vector<vector<char>> VVc;
typedef vector<vector<vector<ll>>> VVV;
typedef vector<vector<vector<vector<ll>>>> VVVV;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define PER(i, a, b) for(ll i=(a); i>=(b); i--)
#define rep(i, n) REP(i, 0, n)
#define per(i, n) PER(i, n, 0)
const ll INF = 1e18+18;
const ll MOD = 1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl;
#define ALL(v) v.begin(), v.end()
#define rALL(v) v.rbegin(), v.rend()
#define pb(x) push_back(x)
#define mp(a, b) make_pair(a,b)
#define Each(a, b) for(auto &a :b)
#define rEach(i, mp) for (auto i = mp.rbegin(); i != mp.rend(); ++i)
#ifdef LOCAL
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbgmap(mp) cerr << #mp << ":"<<endl; for (auto i = mp.begin(); i != mp.end(); ++i) { cerr << i->first <<":"<<i->second << endl;}
#define dbgset(st) cerr << #st << ":"<<endl; for (auto i = st.begin(); i != st.end(); ++i) { cerr << *i <<" ";}cerr<<endl;
#define dbgarr(n, m, arr) rep(i,n){rep(j,m){cerr<<arr[i][j]<<" ";}cerr<<endl;}
#define dbgdp(n, arr) rep(i,n){cerr<<arr[i]<<" ";}cerr<<endl;
#else
#define dbg(...)
#define dbgmap(...)
#define dbgset(...)
#define dbgarr(...)
#define dbgdp(...)
#endif
#define out(a) cout<<a<<endl
#define out2(a, b) cout<<a<<" "<<b<<endl
#define vout(v) rep(i,v.size()){cout<<v[i]<<" ";}cout<<endl
#define Uniq(v) v.erase(unique(v.begin(), v.end()), v.end())
#define fi first
#define se second

template<class T>
bool chmax(T &a, const T &b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmin(T &a, const T &b) {
    if(b < a) {
        a = b;
        return true;
    }
    return false;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

template<typename T>
istream &operator>>(istream &i, vector<T> &v) {
    rep(j, v.size())i>>v[j];
    return i;
}

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len = v.size();
    for(int i = 0; i < len; ++i) {
        s<<v[i];
        if(i < len-1) s<<"	";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len = vv.size();
    for(int i = 0; i < len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}

int solve() {
    ll n;
    cin>>n;
    Vec x(n);
    Vec y(n);
    Vec p(n);
    rep(i, n) {
        cin>>x[i]>>y[i]>>p[i];
    }

    VV g(n+1);
    rep(bit, 1LL<<n) {
        ll pc = __builtin_popcount(bit);
        g[pc].pb(bit);
    }

    VVV G(n+1,VV(n+1));
    rep(k,n+1){
        ll mx = 1LL<<k;
        rep(bit, mx){
            ll pc = __builtin_popcount(bit);
            G[k][pc].pb(bit);
        }
    }

    VV scoreX(n,Vec(1LL<<n));
    VV scoreY(n,Vec(1LL<<n));

    rep(i,n){
        rep(bit, 1LL<<n){
            ll sx = INF;
            ll sy = INF;
            Vec xline;
            xline.pb(0);
            Vec yline;
            yline.pb(0);
            rep(j,n){
                if(bit & (1LL<<j)){
                    xline.pb(x[j]);
                    yline.pb(y[j]);
                }
            }
            Each(X, xline) {
                chmin(sx, abs(x[i]-X)*p[i]);
            }
            Each(Y, yline) {
                chmin(sy, abs(y[i]-Y)*p[i]);
            }
            scoreX[i][bit] = sx;
            scoreY[i][bit] = sy;
        }
    }

    rep(lc, n+1) {
        ll score = INF;
        rep(xlc, lc+1) {
            ll ylc = lc-xlc;

            //x組とy組に分ける
            Vec &xg = g[xlc];

            ll ymax = n-xlc;
            Vec &yg = G[ymax][ylc];
            //dbg(mp(xlc,ylc));
            Each(xbit, xg) {
                Vec cand;
                rep(i, n) {
                    if(!(xbit & (1LL<<i))) {
                        cand.pb(i);
                    }
                }

                ll csz = cand.size();
                Each(bit, yg) {
                    ll ybit = 0;
                    rep(i, csz) {
                        if(bit & (1LL<<i)) {
                            ybit += 1LL<<cand[i];
                        }
                    }
                    //dbg(mp(xbit,ybit));
                    ll sum = 0;
                    rep(i, n) {
                        ll tmp = INF;
                        chmin(tmp, scoreX[i][xbit]);
                        chmin(tmp, scoreY[i][ybit]);
                        sum += tmp;
                    }
                    chmin(score, sum);
                }
            }
        }
        out(score);
    }

    return 0;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<std::setprecision(10);
//    ll T;
//    cin>>T;
//    while(T--)
    solve();
}