#include <iostream>
#include <random>
#include <chrono>
#include <algorithm>
#define rep(i, x) for(ll i = 0; i < x; i++)
using namespace std;
typedef long long ll;
const ll MOD = (ll)1e9+7;
const ll N = 5e5+5, inf = 1e18+5000;
ll add(ll x, ll y) { x += y; if (x >= MOD) return x - MOD; return x; }
ll sub(ll x, ll y) { x -= y; if (x < 0) return x + MOD; return x; }
ll mult(ll x, ll y) { return (x * y) % MOD; }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  
  int n; cin >> n;
  vector<ll> x(n), y(n), p(n);
  rep (i, n) {
    cin >> x[i]; cin >> y[i]; cin >> p[i];
  }

  vector<vector<ll>> horizontals(n, vector<ll>(1<<n)), verticals(n, vector<ll>(1<<n));
  for (int mask = 0; mask < 1<<n; mask++) {
    for (int i = 0; i < n; i++) if (!(mask & (1<<i))) {
      ll best_x = abs(x[i]), best_y = abs(y[i]);
      for (int j = 0; j < n; j++) if (mask & (1<<j)) {
        best_x = min(best_x, abs(x[i] - x[j]));
        best_y = min(best_y, abs(y[i] - y[j]));
      }
      horizontals[i][mask] = best_y;
      verticals[i][mask] = best_x;
    }
  }

  vector<ll> ans(n+1, inf);
  for (int mask = 0; mask < 1<<n; mask++) {
    for (int subm = mask; ; subm = (subm-1)&mask) {
      ll score = 0;
      for (int i = 0; i < n; i++) if (!(mask & (1<<i)))
        score += min(horizontals[i][subm], verticals[i][mask ^ subm]) * p[i];
      ans[__builtin_popcount(mask)] = min(ans[__builtin_popcount(mask)], score);
      if (subm == 0)  break;
    }
  }

  for (int k = 0; k <= n; k++) cout << ans[k] << endl;

  return 0;
}
### 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 <iostream>
#include <random>
#include <chrono>
#include <algorithm>
#define rep(i, x) for(ll i = 0; i < x; i++)
using namespace std;
typedef long long ll;
const ll MOD = (ll)1e9+7;
const ll N = 5e5+5, inf = 1e18+5000;
ll add(ll x, ll y) { x += y; if (x >= MOD) return x - MOD; return x; }
ll sub(ll x, ll y) { x -= y; if (x < 0) return x + MOD; return x; }
ll mult(ll x, ll y) { return (x * y) % MOD; }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  
  int n; cin >> n;
  vector<ll> x(n), y(n), p(n);
  rep (i, n) {
    cin >> x[i]; cin >> y[i]; cin >> p[i];
  }

  vector<vector<ll>> horizontals(n, vector<ll>(1<<n)), verticals(n, vector<ll>(1<<n));
  for (int mask = 0; mask < 1<<n; mask++) {
    for (int i = 0; i < n; i++) if (!(mask & (1<<i))) {
      ll best_x = abs(x[i]), best_y = abs(y[i]);
      for (int j = 0; j < n; j++) if (mask & (1<<j)) {
        best_x = min(best_x, abs(x[i] - x[j]));
        best_y = min(best_y, abs(y[i] - y[j]));
      }
      horizontals[i][mask] = best_y;
      verticals[i][mask] = best_x;
    }
  }

  vector<ll> ans(n+1, inf);
  for (int mask = 0; mask < 1<<n; mask++) {
    for (int subm = mask; ; subm = (subm-1)&mask) {
      ll score = 0;
      for (int i = 0; i < n; i++) if (!(mask & (1<<i)))
        score += min(horizontals[i][subm], verticals[i][mask ^ subm]) * p[i];
      ans[__builtin_popcount(mask)] = min(ans[__builtin_popcount(mask)], score);
      if (subm == 0)  break;
    }
  }

  for (int k = 0; k <= n; k++) cout << ans[k] << endl;

  return 0;
}
```