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;

template<typename T>
inline void remin(T& a, T b) { a = min(a, b); }

vector<long long> solve(int n, const vector<int>& x, const vector<int>& y, const vector<int>& p)
{
    long long inf = (1ll << 62);
    vector<long long> ans(n + 1, inf);
    vector< vector<int> > xsel(1 << n, vector<int>(n));
    vector< vector<int> > ysel(1 << n, vector<int>(n));
    
    for(int i = 0; i < 1 << n; ++i) {
        for(int j = 0; j < n; ++j) {
            xsel[i][j] = abs(x[j]);
            ysel[i][j] = abs(y[j]);
            for(int k = 0; k < n; ++k) {
                if((i >> k) & 1) {
                    remin(xsel[i][j], abs(x[j] - x[k]));
                    remin(ysel[i][j], abs(y[j] - y[k]));
                }
            }
        }
    }
    // Ate agora pagamos O(2^n * n^2), agora vem a parte tricky que domina a complexidade
    for(int i = 0; i < (1 << n); ++i) {
        int cnt = __builtin_popcount(i);
        for(int j = i; j >= 0; --j) {
            j &= i;
            long long cost = 0;
            for(int k = 0; k < n; ++k) {
                if(!((i >> k) & 1)) {
                    cost += 1LL * min(xsel[j][k], ysel[i - j][k]) * p[k];
                }
            }
            remin(ans[cnt], cost);
        }
    }
    return ans;
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int n;
    cin >> n;
    vector<int> x(n), y(n), p(n);
    
    for(int i = 0; i < n; ++i)
    {
        cin >> x[i] >> y[i] >> p[i];
    }
    
    auto ans = solve(n, x, y, p);
    for(int i = 0; i <= n; ++i) cout << ans[i] << '\n';
    return 0;

}