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: Accepted)

#include <cmath>
#include <random>
#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
mt19937 mt(2007111352);
int rand_rng(int l, int r) {
	uniform_int_distribution<int> p(l, r - 1);
	return p(mt);
}
double randouble() {
	uniform_real_distribution<double> p(0.0, 1.0);
	return p(mt);
}
long long calc(int N, vector<int> X, vector<int> Y, vector<int> P, vector<int> xrail, vector<int> yrail) {
	long long ans = 0;
	for (int i = 0; i < N; ++i) {
		int mindist = (1 << 30);
		for (int j : xrail) {
			mindist = min(mindist, abs(X[i] - j));
		}
		for (int j : yrail) {
			mindist = min(mindist, abs(Y[i] - j));
		}
		ans += 1LL * mindist * P[i];
	}
	return ans;
}
int main() {
	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];
	}
	vector<int> xinc, yinc;
	for (int i = 0; i < N; ++i) {
		if (X[i] != 0) xinc.push_back(X[i]);
		if (Y[i] != 0) yinc.push_back(Y[i]);
	}
	sort(xinc.begin(), xinc.end()); xinc.erase(unique(xinc.begin(), xinc.end()), xinc.end());
	sort(yinc.begin(), yinc.end()); yinc.erase(unique(yinc.begin(), yinc.end()), yinc.end());
	int candx = xinc.size();
	int candy = yinc.size();
	vector<long long> ans(N + 1);
	int limit = min(candx, candy);
	vector<bool> flag(candx + candy, false);
	function<long long()> get_score = [&]() {
		vector<int> xrail = { 0 }, yrail = { 0 };
		for (int i = 0; i < candx + candy; ++i) {
			if (flag[i]) {
				if (i < candx) xrail.push_back(xinc[i]);
				else yrail.push_back(yinc[i - candx]);
			}
		}
		return calc(N, X, Y, P, xrail, yrail);
	};
	ans[0] = get_score();
	for (int i = 1; i < limit; ++i) {
		int u = -1;
		do {
			u = rand_rng(0, candx + candy);
		} while (flag[u]);
		flag[u] = true;
		// --- hill climbing! --- //
		long long current_score = get_score();
		long long best_score = current_score;
		int repeat = 100000;
		for (int j = 1; j <= repeat; ++j) {
			int u1 = -1, u2 = -1;
			do {
				u1 = rand_rng(0, candx + candy);
				u2 = rand_rng(0, candx + candy);
			} while (!flag[u1] || flag[u2]);
			flag[u1] = false; // true -> false
			flag[u2] = true; // false -> true
			double temperature = double(max(repeat - (j % (repeat / 4)) * 5, 1)) / repeat * ans[i - 1] * 2.0;
			long long new_score = get_score();
			if (randouble() < exp((current_score - new_score) / temperature)) {
				if (best_score > new_score) {
					cerr << "Repeat #" << j << ": Score " << best_score << " -> " << new_score << endl;
					best_score = new_score;
				}
				current_score = new_score;
			}
			else {
				// revert
				flag[u1] = true;
				flag[u2] = false;
			}
		}
		ans[i] = best_score;
	}
	for (int i = 0; i <= N; ++i) {
		cout << ans[i] << endl;
	}
	return 0;
}