#include<bits/stdc++.h>
using namespace std;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i,n) for(UL i=0; i<(n); i++)

int N;
LL P[15][2];
LL Z[15];

LL XD[1<<15][15];
LL YD[1<<15][15];

vector<UL> MX;
vector<UL> M;

UL popcount(UL x){
 x = (x&0x55555555) + ((x>>1)&0x55555555);
 x = (x&0x33333333) + ((x>>2)&0x33333333);
 x = (x&0x0f0f0f0f) + ((x>>4)&0x0f0f0f0f);
 x = (x&0x00ff00ff) + ((x>>8)&0x00ff00ff);
 x = (x&0x0000ffff) + ((x>>16)&0x0000ffff);
 return x;
}

LL ans[16];

int main() {
 cin>>N;
 rep(i,N) cin>>P[i][0]>>P[i][1]>>Z[i];

 rep(i,N) XD[0][i]=abs(P[i][0]);
 rep(i,N) YD[0][i]=abs(P[i][1]);
 rep(i,15)rep(j,15) rep(m,1<<j){
  int p=m|(1<<j);
  XD[p][i]=min(XD[m][i],abs(P[j][0]-P[i][0]));
  YD[p][i]=min(YD[m][i],abs(P[j][1]-P[i][1]));
 }
 rep(i,15) rep(p,1<<15){ XD[p][i] *= Z[i]; YD[p][i] *= Z[i]; }

 rep(i,1<<10) if(!(i&(i<<5)))
  MX.push_back((i>>5<<15)|(i&0x1f));
 for(UL x:MX) for(UL y:MX) for(UL z:MX)
  M.push_back((x<<10)|(y<<5)|z);

 rep(i,16) ans[i]=1000000000000000ll;

 for(UL m:M){
  int pops = popcount(m);
  UL xm = m>>15;
  UL ym = m&0x7fff;
  LL tmp=0;
  rep(j,15) tmp += min(XD[xm][j],YD[ym][j]);
  ans[pops] = min(ans[pops],tmp);
 }

 rep(i,N+1) cout<<ans[i]<<endl;

 return 0;
}


### 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
#include<bits/stdc++.h>
using namespace std;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i,n) for(UL i=0; i<(n); i++)

int N;
LL P[15][2];
LL Z[15];

LL XD[1<<15][15];
LL YD[1<<15][15];

vector<UL> MX;
vector<UL> M;

UL popcount(UL x){
 x = (x&0x55555555) + ((x>>1)&0x55555555);
 x = (x&0x33333333) + ((x>>2)&0x33333333);
 x = (x&0x0f0f0f0f) + ((x>>4)&0x0f0f0f0f);
 x = (x&0x00ff00ff) + ((x>>8)&0x00ff00ff);
 x = (x&0x0000ffff) + ((x>>16)&0x0000ffff);
 return x;
}

LL ans[16];

int main() {
 cin>>N;
 rep(i,N) cin>>P[i][0]>>P[i][1]>>Z[i];

 rep(i,N) XD[0][i]=abs(P[i][0]);
 rep(i,N) YD[0][i]=abs(P[i][1]);
 rep(i,15)rep(j,15) rep(m,1<<j){
  int p=m|(1<<j);
  XD[p][i]=min(XD[m][i],abs(P[j][0]-P[i][0]));
  YD[p][i]=min(YD[m][i],abs(P[j][1]-P[i][1]));
 }
 rep(i,15) rep(p,1<<15){ XD[p][i] *= Z[i]; YD[p][i] *= Z[i]; }

 rep(i,1<<10) if(!(i&(i<<5)))
  MX.push_back((i>>5<<15)|(i&0x1f));
 for(UL x:MX) for(UL y:MX) for(UL z:MX)
  M.push_back((x<<10)|(y<<5)|z);

 rep(i,16) ans[i]=1000000000000000ll;

 for(UL m:M){
  int pops = popcount(m);
  UL xm = m>>15;
  UL ym = m&0x7fff;
  LL tmp=0;
  rep(j,15) tmp += min(XD[xm][j],YD[ym][j]);
  ans[pops] = min(ans[pops],tmp);
 }

 rep(i,N+1) cout<<ans[i]<<endl;

 return 0;
}


```