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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin>>n;
    int n2=1,n3=1;
    for(int i=0;i<n;i++)n2*=2,n3*=3;
    int x[n],y[n],p[n];
    for(int i=0;i<n;i++)cin>>x[i]>>y[i]>>p[i];
    long long X[n2][n],Y[n2][n];
    for(int i=0;i<n2;i++){
        vector<long long>vx={-100000,0,100000};
        vector<long long>vy={-100000,0,100000};
        int I=i;
        for(int j=0;j<n;j++){
            if(I%2){
                vx.push_back(x[j]);
                vy.push_back(y[j]);
            }
            I/=2;
        }
        sort(vx.begin(),vx.end());
        sort(vy.begin(),vy.end());
        for(int j=0;j<n;j++){
            X[i][j]=min(*upper_bound(vx.begin(),vx.end(),x[j])-x[j],
                       x[j]-*(upper_bound(vx.begin(),vx.end(),x[j])-1))*p[j];
            Y[i][j]=min(*upper_bound(vy.begin(),vy.end(),y[j])-y[j],
                       y[j]-*(upper_bound(vy.begin(),vy.end(),y[j])-1))*p[j];
        }
    }
    vector<long long>ans(n+1,1000000000000000);
    for(int i=0;i<n3;i++){
        int I=i,P=1,build=0,px=0,py=0;
        for(int j=0;j<n;j++){
            if(I%3){
                build++;
                if(I%3==1)px+=P;
                else py+=P;
            }
            P*=2;
            I/=3;
        }
        long long cost=0;
        for(int j=0;j<n;j++)cost+=min(X[px][j],Y[py][j]);
        ans[build]=min(ans[build],cost);
    }
    for(int i=0;i<=n;i++)cout<<ans[i]<<endl;
    return 0;
}