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 <list>
#include <set>
#include <queue>
#include <map>
#include <algorithm>
#include <functional>
#include <string>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <tuple>
using namespace std;

#define REP(i,p,n) for(ll i=p;i<ll(n);++i)
#define rep(i,n) REP(i,0,n)

using ll = long long int;
using pll = pair<ll, ll>;

const ll INF = 1LL << 62;
const ll MAXN = 16;

ll N;

ll X[MAXN];
ll Y[MAXN];
ll P[MAXN];

ll CX[MAXN][1 << MAXN];
ll CY[MAXN][1 << MAXN];

ll calcCost(int xmask, int ymask)
{
    ll cost = 0;
    rep (i, N) { cost += min(CX[i][xmask], CY[i][ymask]) * P[i]; }

    return cost;
}

ll dfs(int index, int last, int xmask, int ymask)
{
    if (index + last > N) { return INF; }
    if (index == N || last == 0) { return calcCost(xmask, ymask); }

    const auto a = dfs(index + 1, last, xmask, ymask);
    const auto b = dfs(index + 1, last - 1, xmask | (1<<index), ymask);
    const auto c = dfs(index + 1, last - 1, xmask, ymask | (1<<index));

    return min(a, min(b, c));
}

int main()
{
    cin >> N;
    rep (i, N) { cin >> X[i] >> Y[i] >> P[i]; }


    rep (i, N) rep (j, 1<<N)
    {
        CX[i][j] = abs(X[i]);
        CY[i][j] = abs(Y[i]);

        rep (k, N)
        {
            if ((j>>k)&1) { CX[i][j] = min(CX[i][j], abs(X[i] - X[k])); }
            if ((j>>k)&1) { CY[i][j] = min(CY[i][j], abs(Y[i] - Y[k])); }
        }
    }


    rep (i, N + 1) { cout << max(0LL, dfs(0, i, 0, 0)) << endl; }

    return 0;
}