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 <iostream>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <string>
#include <sstream>
#include <algorithm>
#include <random>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <cassert>
#include <climits>
#include <bitset>
#include <functional>
#include <iomanip>
#include <random>

#define FOR_LT(i, beg, end) for (int i = (int)(beg); i < (int)(end); i++)
#define FOR_LE(i, beg, end) for (int i = (int)(beg); i <= (int)(end); i++)
#define FOR_DW(i, beg, end) for (int i = (int)(beg); (int)(end) <= i; i--)
#define REP(n)              for (int repeat_index = 0; repeat_index < (int)n; repeat_index++)

double E = 2.71828182845904523536028747;

using namespace std;

struct P {
	int64_t x;
	int64_t y;
	int64_t p;
};

int64_t evaluate(const vector<P>& ps, const vector<int64_t> xl, const vector<int64_t> yl, const vector<bool> sel)
{
	int64_t sum = 0;

	vector<int64_t> cxl;
	vector<int64_t> cyl;
	FOR_LT(i, 0, sel.size()) {
		if (sel[i]) {
			if (i < xl.size()) cxl.push_back(xl[i]);
			else               cyl.push_back(yl[i - xl.size()]);
		}
	}

	for (auto& p : ps) {
		int64_t minv = min(abs(p.x), abs(p.y)) * p.p;
		for (auto& x : cxl) {
			minv = min(minv, abs(p.x - x) * p.p);
		}
		for (auto& y : cyl) {
			minv = min(minv, abs(p.y - y) * p.p);
		}
		sum += minv;
	}
	return sum;
}

int64_t annealing(vector<P>& ps, vector<int64_t>& xl, vector<int64_t>& yl, int n)
{
	if (xl.size() + yl.size() <= n) return 0;
	

	int nl = xl.size() + yl.size();
	vector<bool> sel(nl);

	static random_device rnd;
	static mt19937 mt(rnd());

	uniform_int_distribution<>        sell(0, nl - 1);
	uniform_int_distribution<>        selk(0, n - 1);
	uniform_int_distribution<>        selnk(0, nl - n - 1);
	uniform_real_distribution<double> selr(0, 1.0);

	FOR_LT(i, 0, n) {
		while (true) {
			int l = sell(mt);
			if (sel[l]) continue;
			sel[l] = true;
			break;
		}
	}

	int64_t ans  = evaluate(ps, xl, yl, sel);
	int64_t cans = ans;
	double  t    = ans;
	double  lt   = t;
	int repeat = 100000;
	REP(repeat) {
		//cout << ans << " " << cans << " " << t << endl;
		int  k = selk(mt);
		int nk = selnk(mt);

		int kidx = 0;
		{
			for (int i = 0; i < k; i++) {
				while (!sel[kidx]) kidx++;
				kidx++;
			}
			while (!sel[kidx]) kidx++;
		}

		int nkidx = 0;
		{
			for (int i = 0; i < nk; i++) {
				while (sel[nkidx]) nkidx++;
				nkidx++;
			}
			while (sel[nkidx]) nkidx++;
		}

		sel[kidx]  = false;
		sel[nkidx] = true;
		int64_t nans = evaluate(ps, xl, yl, sel);
		if (nans < cans) {
			cans = nans;
			ans = min(ans, cans);
		}
		else {
			if (selr(mt) < exp((cans - nans) / t)) {
				cans = nans;
			}
			else {
				sel[kidx] = true;
				sel[nkidx] = false;
			}
		}

		lt = 0.9 * lt + 0.1 * (cans - ans);
		if (lt < 1E-10) break;
		t = (repeat - repeat_index)/ repeat * lt;
	}
	
	return ans;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout << fixed << setprecision(20);

	int n; cin >> n;

	vector<P> ps(n);
	FOR_LT(i, 0, n) {
		auto& p = ps[i];
		cin >> p.x >> p.y >> p.p;
	}

	vector<int64_t> xl(n);
	FOR_LT(i, 0, n) {
		xl[i] = ps[i].x;
	}
	sort(xl.begin(), xl.end());
	xl.erase(unique(xl.begin(), xl.end()), xl.end());

	vector<int64_t> yl(n);
	FOR_LT(i, 0, n) {
		yl[i] = ps[i].y;
	}
	sort(yl.begin(), yl.end());
	yl.erase(unique(yl.begin(), yl.end()), yl.end());

	cout << evaluate(ps, xl, yl, vector<bool>()) << endl;
	FOR_LE(i, 1, n) {
		int64_t ans = INT64_MAX;
		for (int k = 0; k < 32; k++) {
			ans = min(ans, annealing(ps, xl, yl, i));
		}
		cout << ans << endl;
	}

	return 0;
}