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<iostream>
#include<vector>
#include<algorithm>
using namespace std;

typedef long long ll;

vector<int> x, y, p;
vector<vector<int>> a, b;
vector<ll> ans;
int n;

int bitx = 0, bity = 0, pop = 0;

void dfs(int ind){
    if(ind == n){
        ll tmp = 0;
        for(int i = 0; i < n; i++){
            int diff = min({abs(x[i]), abs(y[i]), a[i][bitx], b[i][bity]});
            tmp += (ll)diff * p[i];
        }
        ans[pop] = min(ans[pop], tmp);
        return;
    }
    bitx ^= 1<<ind; pop++;
    dfs(ind+1);
    bitx ^= 1<<ind; 
    bity ^= 1<<ind;
    dfs(ind+1);
    bity ^= 1<<ind; pop--;
    dfs(ind+1);
}

int main(){
    cin >> n;
    x.resize(n);
    y.resize(n);
    p.resize(n);
    ans.resize(n+1, 1ll<<60);
    a.resize(n, vector<int>(1<<n, 30000));
    b.resize(n, vector<int>(1<<n, 30000));
    for(int i = 0; i < n; i++)  cin >> x[i] >> y[i] >> p[i];
    for(int i = 0; i < n; i++){
        for(int j = 0; j < 1<<n; j++){
            for(int k = 0; k < n; k++){
                if(j>>k&1)  a[i][j] = min(a[i][j], abs(x[k]-x[i])), b[i][j] = min(b[i][j], abs(y[k]-y[i]));
            }
        }
    }
    dfs(0);
    for(int i = 0; i < n+1; i++)  cout << ans[i] << endl;
    return 0;
}