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;
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define per1(i,n) for(int i=n;i>0;i--)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll inf=1e18;
vec ans(16,inf);
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    vec X(n),Y(n),P(n);
    rep(i,n) cin>>X[i]>>Y[i]>>P[i];
    vec x[15],y[15];
    rep(i,pow(2,n)){
        vec A;
        ll p=i;
        rep(j,n) A.push_back(p%2),p/=2;
        rep(j,n){
            ll d=min(abs(X[j])*P[j],abs(Y[j])*P[j]);
            rep(k,n){
                if(A[k]) d=min(d,abs(X[j]-X[k])*P[j]);
            }
            x[j].push_back(d);
        }
    }
    rep(i,pow(2,n)){
        vec A;
        ll p=i;
        rep(j,n) A.push_back(p%2),p/=2;
        rep(j,n){
            ll d=min(abs(X[j])*P[j],abs(Y[j])*P[j]);
            rep(k,n){
                if(A[k]) d=min(d,abs(Y[j]-Y[k])*P[j]);
            }
            y[j].push_back(d);
        }
    }
    ll e2[16];
    e2[0]=1;
    rep(i,15) e2[i+1]=e2[i]*2;
    rep(i,pow(3,n)){
        vec A,B;
        ll p=i,a=0,b=0,c=0,d=0;
        rep(j,n){
            if(p%3==1) a+=e2[j],d++;
            if(p%3==2) b+=e2[j],d++;
            p/=3;
        }
        rep(j,n){
            c+=min(x[j][a],y[j][b]);
        }
        ans[d]=min(ans[d],c);
    }
    rep(i,n+1) cout<<ans[i]<<"\n";
}