#include <bits/stdc++.h>
using namespace std;
 
using ll = int64_t;
constexpr ll mod = ll(1e9 + 7);
#define endl '\n'


 
void ComputeAdd(vector<array<ll,3>>& A, vector<ll>& min_dist_n_street, vector<int>& ys, vector<int>& xs) {
  int n_street = xs.size() + ys.size();
  ll all_dist = 0;
  for (int ii = 0; ii < A.size(); ++ii) {
    auto& [x,y,p] = A[ii];
    ll dist = min(abs(x),abs(y));
    for (int xx : xs) dist = min(dist, abs(xx-x));
    for (int yy : ys) dist = min(dist, abs(yy-y));
    all_dist += dist*p;
  }
  min_dist_n_street[n_street] = min(min_dist_n_street[n_street], all_dist);
}


void DFS(int i, vector<array<ll,3>>& A, vector<ll>& min_dist_n_street, vector<int>& ys, vector<int>& xs) {
  if (i == A.size()) return;
  DFS(i+1, A, min_dist_n_street, ys, xs);
  

  xs.push_back(A[i][0]);
  ComputeAdd(A, min_dist_n_street, ys, xs);
  DFS(i+1, A, min_dist_n_street, ys, xs);
  xs.pop_back();

  ys.push_back(A[i][1]);
  ComputeAdd(A, min_dist_n_street, ys, xs);
  DFS(i+1, A, min_dist_n_street, ys, xs);
  ys.pop_back();
}


int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  // cin >> TT;
  ll N;
  cin >> N;
  vector<array<ll,3>> A(N);
  for (auto& aa : A) cin >> aa[0] >> aa[1] >> aa[2];
  vector<ll> min_dist_n_street(N+1, INT64_MAX/2);
  vector<int> xs;
  vector<int> ys;
  ComputeAdd(A, min_dist_n_street, ys, xs);
  DFS(0,  A, min_dist_n_street, ys, xs);

  for (int k = 0; k <= N; ++k) {
    cout << min_dist_n_street[k] << endl;
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

Examples

Input

3
1 2 300
3 3 600
1 4 800


Output

2900
900
0
0


Input

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400


Output

13800
1600
0
0
0
0


Input

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200


Output

26700
13900
3200
1200
0
0
0


Input

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


Output

2576709
1569381
868031
605676
366338
141903
0
0
0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
 
using ll = int64_t;
constexpr ll mod = ll(1e9 + 7);
#define endl '\n'


 
void ComputeAdd(vector<array<ll,3>>& A, vector<ll>& min_dist_n_street, vector<int>& ys, vector<int>& xs) {
  int n_street = xs.size() + ys.size();
  ll all_dist = 0;
  for (int ii = 0; ii < A.size(); ++ii) {
    auto& [x,y,p] = A[ii];
    ll dist = min(abs(x),abs(y));
    for (int xx : xs) dist = min(dist, abs(xx-x));
    for (int yy : ys) dist = min(dist, abs(yy-y));
    all_dist += dist*p;
  }
  min_dist_n_street[n_street] = min(min_dist_n_street[n_street], all_dist);
}


void DFS(int i, vector<array<ll,3>>& A, vector<ll>& min_dist_n_street, vector<int>& ys, vector<int>& xs) {
  if (i == A.size()) return;
  DFS(i+1, A, min_dist_n_street, ys, xs);
  

  xs.push_back(A[i][0]);
  ComputeAdd(A, min_dist_n_street, ys, xs);
  DFS(i+1, A, min_dist_n_street, ys, xs);
  xs.pop_back();

  ys.push_back(A[i][1]);
  ComputeAdd(A, min_dist_n_street, ys, xs);
  DFS(i+1, A, min_dist_n_street, ys, xs);
  ys.pop_back();
}


int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  // cin >> TT;
  ll N;
  cin >> N;
  vector<array<ll,3>> A(N);
  for (auto& aa : A) cin >> aa[0] >> aa[1] >> aa[2];
  vector<ll> min_dist_n_street(N+1, INT64_MAX/2);
  vector<int> xs;
  vector<int> ys;
  ComputeAdd(A, min_dist_n_street, ys, xs);
  DFS(0,  A, min_dist_n_street, ys, xs);

  for (int k = 0; k <= N; ++k) {
    cout << min_dist_n_street[k] << endl;
  }
}

```