#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> p_ll;

template<class T>
void debug(T itr1, T itr2) { auto now = itr1; while(now<itr2) { cout << *now << " "; now++; } cout << endl; }
#define repr(i,from,to) for (ll i=(ll)from; i<(ll)to; i++)
#define all(vec) vec.begin(), vec.end()
#define rep(i,N) repr(i,0,N)
#define per(i,N) for (int i=(int)N-1; i>=0; i--)

constexpr ll LLINF = 1LL<<60-1;

int main() {
  ll N; cin >> N;
  ll X[N], Y[N], P[N]; rep(i,N) cin >> X[i] >> Y[i] >> P[i];
  ll xmin[1<<N][N], ymin[1<<N][N];
  rep(i,1<<N) rep(j,N) {
    xmin[i][j] = abs(X[j]); rep(k,N) if (i&1<<k) xmin[i][j] = min(xmin[i][j], abs(X[k]-X[j]));
    ymin[i][j] = abs(Y[j]); rep(k,N) if (i&1<<k) ymin[i][j] = min(ymin[i][j], abs(Y[k]-Y[j]));
  }
  // rep(i,N) debug(xmin[i],xmin[i]+(1<<N));

  ll result[N+1]; rep(i,N+1) result[i] = LLINF;
  const int pw=pow(3,N);
  rep(i,pw) {
    ll ti = i, dx = 0, dy = 0, cnt = N;
    rep(j,N) {
      const int rem = ti%3;
      if (rem==2) { dx += 1<<j; }
      else if (rem==1) { dy += 1<<j; }
      else cnt--;
      ti /= 3;
    }
    ll sum = 0; rep(j,N) sum += min(xmin[dx][j], ymin[dy][j]) * P[j];
    result[cnt] = min(result[cnt], sum);
  }
  rep(i,N+1) cout << result[i] << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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 <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> p_ll;

template<class T>
void debug(T itr1, T itr2) { auto now = itr1; while(now<itr2) { cout << *now << " "; now++; } cout << endl; }
#define repr(i,from,to) for (ll i=(ll)from; i<(ll)to; i++)
#define all(vec) vec.begin(), vec.end()
#define rep(i,N) repr(i,0,N)
#define per(i,N) for (int i=(int)N-1; i>=0; i--)

constexpr ll LLINF = 1LL<<60-1;

int main() {
  ll N; cin >> N;
  ll X[N], Y[N], P[N]; rep(i,N) cin >> X[i] >> Y[i] >> P[i];
  ll xmin[1<<N][N], ymin[1<<N][N];
  rep(i,1<<N) rep(j,N) {
    xmin[i][j] = abs(X[j]); rep(k,N) if (i&1<<k) xmin[i][j] = min(xmin[i][j], abs(X[k]-X[j]));
    ymin[i][j] = abs(Y[j]); rep(k,N) if (i&1<<k) ymin[i][j] = min(ymin[i][j], abs(Y[k]-Y[j]));
  }
  // rep(i,N) debug(xmin[i],xmin[i]+(1<<N));

  ll result[N+1]; rep(i,N+1) result[i] = LLINF;
  const int pw=pow(3,N);
  rep(i,pw) {
    ll ti = i, dx = 0, dy = 0, cnt = N;
    rep(j,N) {
      const int rem = ti%3;
      if (rem==2) { dx += 1<<j; }
      else if (rem==1) { dy += 1<<j; }
      else cnt--;
      ti /= 3;
    }
    ll sum = 0; rep(j,N) sum += min(xmin[dx][j], ymin[dy][j]) * P[j];
    result[cnt] = min(result[cnt], sum);
  }
  rep(i,N+1) cout << result[i] << endl;
  return 0;
}

```