#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
#define rep(i, begin, n) for (int i = begin; i < n; i++)
#define repe(i, begin, n) for (int i = begin; i <= n; i++)
#define repr(i, begin, n) for (int i = begin; i > begin - n; i--)
#define repre(i, begin, end) for (int i = begin; i >= end; i--)

template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}

const int inf = 1000000007;
const int MOD = 1000000007;
const long long INF = 1000000000000000007;

// -------------------------------------------------------

ll N;
ll X[30], Y[30], P[30], dist[30];
long double eps = 1e-9;
ll ans[30];

void init_dist() {
  rep(i, 0, N) {
    ll d = INF;
    chmin(d, abs(X[i]));
    chmin(d, abs(Y[i]));
    dist[i] = d;
  }
}

void dfs(int cur, int cnt) {
  if (cur >= N) {
    ll score = 0;
    rep(i, 0, N) { score += dist[i] * P[i]; }
    chmin(ans[cnt], score);
    return;
  }

  dfs(cur + 1, cnt);

  ll cdist[30];
  rep(i, 0, N) {
    cdist[i] = dist[i];
    chmin(dist[i], abs(Y[i] - Y[cur]));
  }
  dfs(cur + 1, cnt + 1);

  rep(i, 0, N) { dist[i] = min(cdist[i], abs(X[i] - X[cur])); }
  dfs(cur + 1, cnt + 1);

  rep(i, 0, N) { dist[i] = cdist[i]; }
}

int main() {
  //
  cin >> N;
  rep(i, 0, N) { cin >> X[i] >> Y[i] >> P[i]; }

  init_dist();
  repe(i, 0, N) { ans[i] = INF; }

  dfs(0, 0);

  cout << ans[0] << "\n";
  repe(i, 1, N) {
    chmin(ans[i], ans[i - 1]);
    cout << ans[i] << "\n";
  }
}

### Prompt

Create a solution in cpp for the following 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

### Response

```cpp
#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
#define rep(i, begin, n) for (int i = begin; i < n; i++)
#define repe(i, begin, n) for (int i = begin; i <= n; i++)
#define repr(i, begin, n) for (int i = begin; i > begin - n; i--)
#define repre(i, begin, end) for (int i = begin; i >= end; i--)

template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}

const int inf = 1000000007;
const int MOD = 1000000007;
const long long INF = 1000000000000000007;

// -------------------------------------------------------

ll N;
ll X[30], Y[30], P[30], dist[30];
long double eps = 1e-9;
ll ans[30];

void init_dist() {
  rep(i, 0, N) {
    ll d = INF;
    chmin(d, abs(X[i]));
    chmin(d, abs(Y[i]));
    dist[i] = d;
  }
}

void dfs(int cur, int cnt) {
  if (cur >= N) {
    ll score = 0;
    rep(i, 0, N) { score += dist[i] * P[i]; }
    chmin(ans[cnt], score);
    return;
  }

  dfs(cur + 1, cnt);

  ll cdist[30];
  rep(i, 0, N) {
    cdist[i] = dist[i];
    chmin(dist[i], abs(Y[i] - Y[cur]));
  }
  dfs(cur + 1, cnt + 1);

  rep(i, 0, N) { dist[i] = min(cdist[i], abs(X[i] - X[cur])); }
  dfs(cur + 1, cnt + 1);

  rep(i, 0, N) { dist[i] = cdist[i]; }
}

int main() {
  //
  cin >> N;
  rep(i, 0, N) { cin >> X[i] >> Y[i] >> P[i]; }

  init_dist();
  repe(i, 0, N) { ans[i] = INF; }

  dfs(0, 0);

  cout << ans[0] << "\n";
  repe(i, 1, N) {
    chmin(ans[i], ans[i - 1]);
    cout << ans[i] << "\n";
  }
}

```