Problem: 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

Solution (in CPP):

Solution:
// -----------------------------------
// author   : MatsuTaku
// country  : Japan
// created  : 07/25/20 20:57:10
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

constexpr int INF = 1e9;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int n; cin>>n;
  struct City {
    int x,y,p;
  };
  vector<City> C(n);
  for (int i = 0; i < n; i++) {
    int x,y,p; cin>>x>>y>>p;
    C[i] = {x, y, p};
  }
//  n = 15;
  ll distx[1<<n][n], disty[1<<n][n];
  for (int mask = 0; mask < (1<<n); mask++) {
    for (int i = 0; i < n; i++) {
      int mindistx = abs(C[i].x), mindisty = abs(C[i].y);
      for (int j = 0; j < n; j++) {
        if ((mask & (1<<j)) == 0) continue;
        mindistx = min(mindistx, abs(C[i].x-C[j].x));
        mindisty = min(mindisty, abs(C[i].y-C[j].y));
      }
      distx[mask][i] = (ll)mindistx * C[i].p;
      disty[mask][i] = (ll)mindisty * C[i].p;
    }
  }
  vector<ll> ans(n+1, 1e18);
  for (int mask = 0; mask < (1<<n); mask++) {
    int popcnt = __builtin_popcount(mask);
    for (int xmask = mask; xmask >= 0; xmask=(xmask-1)&mask) {
      int ymask = mask^xmask;
      ll sum = 0;
      for (int i = 0; i < n; i++) {
        sum += min(distx[xmask][i], disty[ymask][i]);
      }
      ans[popcnt] = min(ans[popcnt], sum);
      if (xmask == 0)
        break;
    }
  }
  for (auto a:ans)
    cout << a << endl;

  return 0;
}