// solution for m-solutions2020-E
// author : lynmisakura
#include<bits/stdc++.h>
using namespace std;

using ll = long long;
#define REP(i,n) for(int (i) = 0;(i) < (n);(i)++)

template<class T> bool chmin(T& a,T b){ if(b < a){a = b;return true;}else return false;}
template<class T> bool chmax(T& a,T b){ if(b > a){a = b;return true;}else return false;}

#define maxn 15

int n;
ll x[maxn],y[maxn],p[maxn];

void read(){
  cin >> n;
  REP(i,n)cin >> x[i] >> y[i] >> p[i];
}

ll par[maxn][1 << maxn],ver[maxn][1 << maxn];

void calc_par(int i,int mask){
  ll dist = abs(y[i]);
  REP(j,n){
    if(mask >> j & 1){
      chmin(dist,abs(y[j] - y[i]));
    }
  }
  par[i][mask] = dist * p[i];
}

void calc_ver(int i,int mask){
  ll dist = abs(x[i]);
  REP(j,n){
    if(mask >> j & 1){
      chmin(dist,abs(x[j] - x[i]));
    }
  }
  ver[i][mask] = dist * p[i];
}

void pre(){
  REP(mask,(1 << n)){
    REP(i,n){
      calc_par(i,mask);
      calc_ver(i,mask);
    }
  }
}

ll ans[maxn+1];

void solve(int mask_p,int mask_v){
  int ps = __builtin_popcount(mask_p);
  int pv = __builtin_popcount(mask_v);

  ll res = 0;

  REP(i,n){
    res += min(par[i][mask_p] , ver[i][mask_v]);
  }
  
  chmin(ans[ps + pv] , res);
}

int main(void){
  cin.tie(0);
  ios::sync_with_stdio(false);

  read();
  pre();
  
  REP(i,n+1) ans[i] = (1LL << 55);

  REP(mask_v,1 << n){
    int left = (1<< n) - 1 - mask_v;
    int mask_p = left;
    do{
      solve(mask_p,mask_v);
      mask_p = (mask_p - 1) & left;
    }while(mask_p != left);
  } 

  REP(i,n+1)cout << ans[i] << '\n';

}

### Prompt

Generate a Cpp solution to 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
// solution for m-solutions2020-E
// author : lynmisakura
#include<bits/stdc++.h>
using namespace std;

using ll = long long;
#define REP(i,n) for(int (i) = 0;(i) < (n);(i)++)

template<class T> bool chmin(T& a,T b){ if(b < a){a = b;return true;}else return false;}
template<class T> bool chmax(T& a,T b){ if(b > a){a = b;return true;}else return false;}

#define maxn 15

int n;
ll x[maxn],y[maxn],p[maxn];

void read(){
  cin >> n;
  REP(i,n)cin >> x[i] >> y[i] >> p[i];
}

ll par[maxn][1 << maxn],ver[maxn][1 << maxn];

void calc_par(int i,int mask){
  ll dist = abs(y[i]);
  REP(j,n){
    if(mask >> j & 1){
      chmin(dist,abs(y[j] - y[i]));
    }
  }
  par[i][mask] = dist * p[i];
}

void calc_ver(int i,int mask){
  ll dist = abs(x[i]);
  REP(j,n){
    if(mask >> j & 1){
      chmin(dist,abs(x[j] - x[i]));
    }
  }
  ver[i][mask] = dist * p[i];
}

void pre(){
  REP(mask,(1 << n)){
    REP(i,n){
      calc_par(i,mask);
      calc_ver(i,mask);
    }
  }
}

ll ans[maxn+1];

void solve(int mask_p,int mask_v){
  int ps = __builtin_popcount(mask_p);
  int pv = __builtin_popcount(mask_v);

  ll res = 0;

  REP(i,n){
    res += min(par[i][mask_p] , ver[i][mask_v]);
  }
  
  chmin(ans[ps + pv] , res);
}

int main(void){
  cin.tie(0);
  ios::sync_with_stdio(false);

  read();
  pre();
  
  REP(i,n+1) ans[i] = (1LL << 55);

  REP(mask_v,1 << n){
    int left = (1<< n) - 1 - mask_v;
    int mask_p = left;
    do{
      solve(mask_p,mask_v);
      mask_p = (mask_p - 1) & left;
    }while(mask_p != left);
  } 

  REP(i,n+1)cout << ans[i] << '\n';

}

```