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>

#define double long double
#define ff first
#define ss second
#define endl '\n'
#define ii pair<int, int>
#define mp make_pair
#define mt make_tuple
#define DESYNC                      \
  ios_base::sync_with_stdio(false); \
  cin.tie(0);                       \
  cout.tie(0)
#define pb push_back
#define vi vector<int>
#define vii vector<ii>
#define all(x) x.begin(), x.end()
#define EPS 1e-9
#define INF 1e18
#define ROOT 1
#define M 1000000007
#define curtime chrono::steady_clock::now().time_since_epoch().count
#define rep(i, beg, n, s) for (int i = beg; i < n; i += s)
const double PI = acos(-1);

using namespace std;

inline int mod(int n, int m = M) {
  int ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}

int exp(int n, int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  int ax = exp(n, k / 2);
  ax = mod(ax * ax);
  if (k % 2) ax = mod(ax * n);
  return ax;
}

int gcd(int a, int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}

struct cit {
  int x, y, p;
  cit() {}
  cit(int x, int y, int p) : x(x), y(y), p(p) {}
};

long long best[20];
cit v[16];
int n;
vector<int> masks_by_size[20];

//#define MULTIPLE_TEST_CASE
void solution() {
  int n;
  cin >> n;
  vector<cit> v(n);
  for (int i = 0; i < n; i++) cin >> v[i].x >> v[i].y >> v[i].p;
  set<int> sx;
  for (int i = 0; i < n; i++) {
    sx.insert(v[i].x);
  }
  vector<int> vx;
  for (int x : sx) {
    vx.pb(x);
  }
  for (int i = 0; i <= n + 1; i++) best[i] = INF;

  // prepare masks
  for (int msk = 0; msk < (1 << n); msk++) {
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      if ((1 << i) & msk) cnt++;
    }
    masks_by_size[cnt].pb(msk);
  }

  // BRUTE X
  for (int mskx = 0; mskx < (1 << vx.size()); mskx++) {
    // for each submask x brute y
    set<int> selected_x;
    for (int i = 0; i < vx.size(); i++) {
      if ((1 << i) & mskx) selected_x.insert(vx[i]);
    }
    set<int> sy;
    vector<int> vy;

    for (int i = 0; i < n; i++) {
      if (!selected_x.count(v[i].x)) sy.insert(v[i].y);
    }
    for (int y : sy) vy.pb(y);

    int cntx = 0;
    vector<int> xs;
    for (int i = 0; i < vx.size(); i++) {
      if ((1 << i) & mskx) {
        cntx++;
        xs.pb(vx[i]);
      }
    }

    // BRUTE Y
    for (int sz = 0; sz <= n - cntx; sz++) {
      for (int msky : masks_by_size[sz]) {
        if (msky >= (1 << vy.size())) break;

        int cnty = 0;
        vector<int> ys;
        for (int i = 0; i < vy.size(); i++) {
          if ((1 << i) & msky) {
            cnty++;
            ys.pb(vy[i]);
          }
        }
        int cnt = cntx + cnty;

        // cout << "selected " << mskx << "  " << msky << " ";
        // for (int x : xs) cout << x << " ";
        // cout << endl;
        // cout << "// ";
        // for (int y : ys) cout << y << " ";
        // cout << endl;
        long long ans = 0;
        for (int i = 0; i < n; i++) {
          long long cur = INF;
          for (int x : xs) cur = min(cur, 1LL * v[i].p * abs(x - v[i].x));
          for (int y : ys) cur = min(cur, 1LL * v[i].p * abs(y - v[i].y));
          cur = min(cur, 1LL * v[i].p * abs(v[i].x));
          cur = min(cur, 1LL * v[i].p * abs(v[i].y));
          ans += cur;
        }
        best[cnt] = min(best[cnt], ans);
      }
    }
  }
  for (int i = n; i >= 0; i--) {
    if (best[i] == 0) break;
    best[i] = 0;
  }
  for (int i = 0; i <= n; i++) cout << best[i] << endl;
}

int32_t main() {
  DESYNC;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int t = 1;
#ifdef MULTIPLE_TEST_CASE
  cin >> t;
#endif
  while (t--) solution();
}