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:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 17;
const ll INF = 1e16 + 69;
const int M = (1 << 15) + 5;

int n;
ll x[N], y[N], p[N], f[N][M], cost[M];

ll get (vector <pair <ll, ll>> vec) {
  if (vec.empty()) return 0;
  sort(vec.begin(), vec.end());
  ll tot = 0, cur = 0, pt = -1;
  for (auto it : vec) tot += it.second;
  for (auto it : vec) {
    cur += it.second;
    if (cur + cur >= tot) {
      pt = it.first;
      break;
    }
  }
  ll ret = 0;
  for (auto it : vec) ret += it.second * abs(pt - it.first);
  return ret;
}

int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i] >> p[i];
  }
  for (int mask = 0; mask < 1 << n; ++mask) {
    vector <pair <ll, ll>> one, two;
    for (int i = 0; i < n; ++i) if (mask & 1 << i) {
      one.emplace_back(x[i], p[i]);
      two.emplace_back(y[i], p[i]);
    }
    cost[mask] = min(get(one), get(two));
  }
  for (int mask = 0; mask < 1 << n; ++mask) {
    for (int i = 0; i < n; ++i) if (mask & 1 << i) {
      f[0][mask] += p[i] * min(abs(x[i]), abs(y[i]));
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int mask = 1; mask < 1 << n; ++mask) {
      f[i][mask] = INF;
      for (int sub = mask; sub; sub = (sub - 1) & mask) {
        f[i][mask] = min(f[i][mask], cost[sub] + f[i - 1][mask ^ sub]);
      }
    }
  }
  for (int i = 0; i <= n; ++i) cout << f[i][(1 << n) - 1] << '\n';
  return 0;
}