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: Wrong Answer)

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
using ll = long long;
using P = pair<ll,ll>;

int main(){
ios::sync_with_stdio(false);
std::cin.tie(nullptr);
ll n;
cin >> n;
vector<ll> x(n),y(n),p(n);
vector<ll> costs(n,10000000);
set<ll> st_x,st_y;

rep(i,n){
    cin >> x[i] >> y[i] >> p[i];
    st_x.insert(x[i]);
    st_y.insert(y[i]);
}
    ll sum = 0;
    for(int j = 0; j < n; j++){
        sum += min(abs(x[j]) * p[j], abs(y[j]) * p[j]);
    }
    ll k = sum;

set<ll> roads_x,roads_y;
roads_x.insert(0);
roads_y.insert(0);
vector<ll> min_sum(n, 1145141919810364);
for(int i = 0; i < n; i++){
    pair<ll,ll> road_xy({0,-100000});
    
        for(auto v: st_x){
            sum = 0;
            for(int j = 0; j < n; j++){
                if(roads_x.count(x[j]) || roads_y.count(y[j]))
                    sum += 0;
                else
                   sum += abs(v - x[j]) * p[j];
            }
            if(sum < min_sum[i]) {
                road_xy = make_pair(0, v);
                min_sum[i] = sum;
            }
        }
        for(auto v: st_y){
            sum = 0;
            for(int j = 0; j < n; j++){
                if(roads_x.count(x[j]) || roads_y.count(y[j]))
                    sum += 0;
                else
                    sum += abs(v - y[j]) * p[j];
            }
            if(sum < min_sum[i]) {
                road_xy = make_pair(1, v);
                min_sum[i] = sum;
            }
        }
        if(road_xy.first == 0) roads_x.insert(road_xy.second);
        if(road_xy.first == 1) roads_y.insert(road_xy.second);
        ll rec = 0;
        for(int j = 0; j < n; j++){
            if(roads_x.count(x[j]) || roads_y.count(y[j])){}
            else{
                ll con = 10000000000;
                for(auto v: roads_x){
                    con = min(con, abs(v - x[j]) * p[j]);
                }
                for(auto v: roads_y){
                    con = min(con, abs(v - y[j]) * p[j]);
                }
                rec += con;
            }
        }
        costs[i] = rec;
}

for(int i = 0; i < n; i++){
    if(i == 0)
        cout << k << endl;
    cout << costs[i] << endl;
}
}