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)

// Marcin Knapik
#include<bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")

using namespace std;

// #include <ext/pb_ds/assoc_container.hpp> // order_of_key
// #include <ext/pb_ds/tree_policy.hpp> // find by order
// using namespace __gnu_pbds; 
// #define ordered_set tree< int , null_type,less< int >, rb_tree_tag,tree_order_statistics_node_update>  

// const ll MAX_LL = 0xFFFFFFFFFFFFFFFF;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define losuj(a, b) uniform_int_distribution<ll>(a, b)(rng)
 
typedef long double             ld;
typedef long long               ll;
typedef pair<int, int>          ii;
typedef vector<int>             vi;
typedef vector<ll>              vll;
typedef vector<vll>             vvll;
typedef vector<ii>              vii;
typedef vector<vi>              vvi;
typedef vector<vii>             vvii;
#define pll pair<long long, long long>
#define vpll vector<pll> 
#define ull unsigned long long

#define sz(a)                   (int)(a).size()
#define pb                      push_back
#define all(c)                  (c).begin(), (c).end()
#define rall(c)                 (c).rbegin(), (c).rend()
#define FOR(i, a)               for (int i = 0; i < (a); i++)
#define f                       first
#define s                       second

template <class T> inline bool setmin(T &a, T b){if (a > b)return a = b, 1;return 0;}
template <class T> inline bool setmax(T &a, T b){if (a < b)return a = b, 1;return 0;}

template<class T> inline T fast(T a,T b,T mod) {ll res = 1; while(b){if(b&1) res = (res*a)%mod;a = (a*a)%mod;b >>= 1;}return res;}
template<class T> istream &operator>>(istream &os, vector<T> &container){for (auto &u : container)os >> u;return os;}
template<class T> ostream &operator<<(ostream &os, const vector<T> &container){for (auto &u : container)os << u << " ";return os;}
template<class T> inline T gcd(T a, T b) { while (b)swap(a %= b, b); return abs(a); }

const ll INF = 1e9 + 7;
const ll mod = 998244353;
const ll BIG_INF = 1e18 + 7;
const ll N = 1e6+7;
const ll T = 1<<20;
const ld inf = 1e18;

ll n, m, q, k, r;
ll X[15];
ll Y[15];
ll P[15];
vll ans(15, BIG_INF);

int stan[15];

ll koszt(){
    ll ret = 0;
    for(int i = 0; i < n; i++){
        ll mini = min(abs(X[i]), abs(Y[i]));

        for(int j = 0; j < n; j++){
            if(stan[j] == 1)
                mini = min(mini, abs(X[i] - X[j]));
            if(stan[j] == 2)
                mini = min(mini, abs(Y[i] - Y[j]));
        }
        ret += P[i] * mini;
    }
    return ret;
}

int ile_bitow(){
    int ret = 0;
    for(int i = 0; i < n; i++)
        ret += stan[i] > 0;
    return ret;
}

void backtrack(int poz){
    if(poz == n){
        setmin(ans[ile_bitow()], koszt());
        return;
    }

    FOR(i, 3){
        stan[poz] = i;
        backtrack(poz + 1);
    }
}

void solve(){
    cin >> n;
    FOR(i, n)
        cin >> X[i] >> Y[i] >> P[i];
    backtrack(0);
    FOR(i, n + 1)
        cout << ans[i] << '\n';
}

signed main(){
  ios::sync_with_stdio(0);
  cin.tie(0);

  q = 1;
  // cin >> q;

  for(int i = 1; i <= q; i++)
    solve();
}