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;
#define int long long
#define FOR(i, a, b) for(int i=(a);i<(b);i++)
#define RFOR(i, a, b) for(int i=(b-1);i>=(a);i--)
#define REP(i, n) for(int i=0; i<(n); i++)
#define RREP(i, n) for(int i=(n-1); i>=0; i--)
#define REP1(i, n) for(int i=1; i<=(n); i++)
#define RREP1(i, n) for(int i=(n); i>=1; i--)
#define ALL(a) (a).begin(),(a).end()
#define UNIQUE_SORT(l) sort(ALL(l)); l.erase(unique(ALL(l)), l.end());
#define CONTAIN(a, b) find(ALL(a), (b)) != (a).end()
#define out(...) printf(__VA_ARGS__)
#if DEBUG
#define debug(...) printf(__VA_ARGS__)
#else
#define debug(...) /* ... */
#endif
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;}

void solve();
signed main()
{
#if DEBUG
    std::ifstream in("input.txt");
    std::cin.rdbuf(in.rdbuf());
#endif
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}
/*================================*/
#if DEBUG
#define SIZE 100
#else
#define SIZE 223450
#endif

int N,Q,K;
int A[SIZE];

const int MAX = 5e5;
const int MOD = 1e9+7;

long long fac[MAX], finv[MAX], inv[MAX];

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    if (!fac[n]) COMinit();
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int P(int n, int k) {
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    if (!fac[n]) COMinit();
    return (fac[n] * finv[n-k]) % MOD;
}

struct Item {
    int x,y,p;
};
vector<Item> V;

int calc(int i, int j) {
    // iに含まれる駅を含む線
    // jのbitが1なら縦,0なら横
    set<int> xs={0}, ys={0};
    int jcnt = 0;
    REP(k,N) if (i>>k&1) {
        if (j>>jcnt&1) {
            // 縦
            xs.insert(V[k].x);
        } else {
            // 横
            ys.insert(V[k].y);
        }
        jcnt++;
    }
    // コスト
    int ret = 0;
    REP(k,N) {
        int mi = LONG_LONG_MAX;
        int x = V[k].x, y = V[k].y, p = V[k].p;
        auto itx = xs.lower_bound(x);
        if (itx!=xs.end()) chmin(mi, *itx-x);
        if (itx!=xs.begin()) chmin(mi, x-*(--itx));
        auto ity = ys.lower_bound(y);
        if (ity!=ys.end()) chmin(mi, *ity-y);
        if (ity!=ys.begin()) chmin(mi, y-*(--ity));
        ret += mi * p;
    }
    return ret;
}

void solve() {
    cin>>N;
    V.resize(N);
    REP(i,N) {
        cin>>V[i].x>>V[i].y>>V[i].p;
    }
    vector<int> ans(N+1,LONG_LONG_MAX/2);
    REP(i,1LL<<N) {
        int b = (int)__builtin_popcount(i);
        int ret = LONG_LONG_MAX/2;
        REP(j,1LL<<b) {
            chmin(ret, calc(i,j));
        }
        chmin(ans[b], ret);
    }
    
    REP(i,N+1) {
        cout << ans[i] << endl;
    }
}