Problem:
Score: 500 points

Problem Statement

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.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

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

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
 
#define $(x) (int)((x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define endl '\n'
#define x first
#define y second
#define int long long
#define double long double
#define pii pair<int, int>
#define pb push_back
#define vec vector
#define beg begin
#define dbg(x) cerr << #x << " = " << x << endl;
 
template<class T> ostream& operator<<(ostream &str, vector<T> &a) {
    for (auto &i : a) {
        str << i << " ";
    }
    return str;
}
 
template<class T> istream& operator>>(istream &str, vector<T> &a) {
    for (auto &i : a) {
        str >> i;
    }
    return str;
}
 
template<class T> ostream& operator<<(ostream &str, pair<T, T> &a) {
    str << a.first << " " << a.second;
    return str;
}
 
template<class T> istream& operator>>(istream &str, pair<T, T> &a) {
    str >> a.first >> a.second;
    return str;
}
 
void solve(); signed main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    srand(time(0));
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cout.setf(ios::fixed); cout.precision(6);
    solve();
    return 0;
}
 
const int MOD = 998244353, MAXN = 5e5 + 10, INF = 1e18 + 10, BASE = 37;
int n;
vec<vec<int>> p;
vec<int> ans;
vec<int> prefix;

void check(int i) {
    if (i == n) {
        int cnt = 0;
        vec<int> h, v;
        for (int i = 0; i < n; ++i) {
            if (prefix[i] >= 1) {
                cnt++;
            }
            if (prefix[i] == 1) {
                h.pb(p[i][1]);
            }
            if (prefix[i] == 2) {
                v.pb(p[i][0]);
            }
        }   
        v.pb(0);
        h.pb(0);
        int cur = 0;
        for (int i = 0; i < n; ++i) {
            int tmp = INF;
            for (int j = 0; j < $(h); ++j) {
                tmp = min(tmp, abs(h[j] - p[i][1]));
            }
            for (int j = 0; j < $(v); ++j) {
                tmp = min(tmp, abs(v[j] - p[i][0]));
            }
            cur += tmp * p[i][2];
        }
        ans[cnt] = min(ans[cnt], cur);
        return;
    }
    else {
        for (int j = 0; j < 3; ++j) {
            prefix.pb(j);
            check(i + 1);
            prefix.pop_back();
        }
    }
}

void solve() {
    cin >> n;
    p.resize(n, vec<int>(3));
    cin >> p;
    ans.resize(n + 1, INF);

    check(0);

    for (int i = 0; i <= n; ++i) {
        cout << ans[i] << endl;
    }
}