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>
#define REP(i, e) for(int (i) = 0; (i) < (e); ++(i))
#define FOR(i, b, e) for(int (i) = (b); (i) < (e); ++(i))
#define ALL(c) (c).begin(), (c).end()
#define PRINT(x) cout << (x) << "\n"
using namespace std;
using ll = long long; using pint = pair<int, int>; using pll = pair<ll, ll>;
template<typename T> auto compare = [](T x, T y) -> bool{return (x < y);};
const long long MOD = 1000000007;

ll N, X[20], Y[20], P[20];
ll cost[1 << 20];
ll dp[1 << 20][20];

void init(){
    REP(st, 1 << N){
        ll z = INT64_MAX / 2;
        REP(i, N){
            ll x = 0, y = 0;
            if((st & (1 << i)) == 0) continue;
            REP(j, N){
                if((st & (1 << j)) == 0) continue;
                x += P[j] * abs(X[i] - X[j]);
                y += P[j] * abs(Y[i] - Y[j]);
            }
            z = min(z, min(x, y));
        }
        cost[st] = z;
    }
}

void solve(){
    init();
    REP(st, 1 << N){
        REP(i, N){
            if(st & (1 << i)) dp[st][0] += P[i] * min(abs(X[i]), abs(Y[i]));
        }
    }
    REP(i, N){
        REP(s, 1 << N){
            ll m = INT64_MAX / 2;
            for(int t = s; t >= 0; t--){
                t &= s;
                m = min(m, dp[s - t][i] + cost[t]);
            }
            dp[s][i + 1] = m;
        }
    }
}

signed main(){
    cin >> N;
    REP(i, N) cin >> X[i] >> Y[i] >> P[i];
    solve();
    REP(i, N + 1) PRINT(dp[(1 << N) - 1][i]);
    return 0;
}