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 ll = long long;
using ull = unsigned long long;
using namespace std;
constexpr int inf = 1<<30;
constexpr int mo = 1e9+7;
constexpr ll infl = 1ll<<60;

int main(){
    cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    
    int n;
    cin>>n;
    vector<tuple<ll,ll,ll>> v;
    for(int i=0;i<n;++i){
        ll x,y,p;
        cin>>x>>y>>p;
        v.push_back(make_tuple(x,y,p));
    }
    vector<vector<ll>> xco(1<<15,vector<ll>(n));
    auto yco = xco;
    for(int mask=0;mask<1<<n;++mask){
        for(int i=0;i<n;++i){
            xco[mask][i] = abs(get<0>(v[i]))*get<2>(v[i]);
            yco[mask][i] = abs(get<1>(v[i]))*get<2>(v[i]);
            for(int j=0;j<n;++j)if(mask&1<<j){
                xco[mask][i] = min(xco[mask][i],abs(get<0>(v[j]) - get<0>(v[i]))*get<2>(v[i]));
                yco[mask][i] = min(yco[mask][i],abs(get<1>(v[j]) - get<1>(v[i]))*get<2>(v[i]));
            }
        }
    }
    vector<ll> ret(n+1,infl);
    for(int xmask=0;xmask<1<<n;++xmask){
        int l = xmask^((1<<n)-1);
        for(int ymask=l;ymask>=0;ymask--){
            ymask &= l;
            ll sum = 0;
            for(int i=0;i<n;++i)sum += min(xco[xmask][i],yco[ymask][i]);
            int num = __builtin_popcount(xmask|ymask);
            ret[num] = min(ret[num],sum);
        }
    }
    ll mi = infl;
    for(int i=0;i<=n;++i){
        mi = min(mi,ret[i]);
        cout << mi << endl;
    }
    return 0;
    
}