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: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<unordered_map>
#include<functional>
#include<map>
#include<iomanip>
#include<limits>
#include<unordered_set>
#include<cmath>
#include <numeric>
#include <array>
#include<utility>
#include <complex>
using namespace std;
#pragma region define
#define M_PI 3.141592653589793238
#define upperbound(v,val) upper_bound(v.begin(),v.end(),val)-v.begin()
#define lowerbound(v,val) lower_bound(v.begin(),v.end(),val)-v.begin()
#define int long long
#define vel vector<long long>
#define vvel vector<vel>
#define rep(i,n) for(int i=0;i<n;i++)
#define sor(v) sort(v.begin(),v.end())
#define mmax(a,b) a=max(a,b)
#define mmin(a,b) a=min(a,b)
#define mkp(a,b) make_pair(a,b)
#define pin pair<int,int>
#define qin pair<pin,int>
#define V vector
#define Endl endl
#define veb vector<bool>
#define fcout cout << fixed << setprecision(15)
#define rev(s) reverse(s.begin(),s.end())
#define lower(h,val) (lower_bound(h.begin(),h.end(),val)-h.begin())
#define upper(h,val) (upper_bound(h.begin(),h.end(),val)-h.begin())
#define vveb V<veb>
#define omajinai cin.tie(0);ios::sync_with_stdio(false);
#define endl "\n"
#define pb push_back
#pragma endregion
#pragma region Inner Class
int root(int x, vel& pa) {
	if (pa[x] == -1) { return x; }
	int ans = root(pa[x], pa); pa[x] = ans;
	return ans;
}
bool mar(int x, int y, vel& pa) {
	x = root(x, pa);
	y = root(y, pa);
	if (x != y) { pa[x] = y; }
	return (x != y);
}
int gcd(int x, int y) {
	if (x < y) { return gcd(y, x); }
	if (y == 0) { return x; }
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	x = abs(x); y = abs(y);
	return x * (y / gcd(x, y));
}
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
vel dijk(V<V<pin>> way, int st, int inf) {
	int n = way.size();
	vel dist(n, inf); dist[st] = 0;
	priority_queue<pin, vector<pin>, greater<pin>> pq;
	pq.push(mkp(0, st));
	veb is_checked(n, false);
	while (!pq.empty()) {
		pin x = pq.top(); pq.pop();
		int pot = x.second;
		if (!is_checked[pot]) {
			is_checked[pot] = true;
			for (auto y : way[pot]) {
				int nex_dist = x.first + y.second;
				int nex_pot = y.first;
				if (dist[nex_pot] > nex_dist) {
					dist[nex_pot] = nex_dist;
					pq.push(mkp(nex_dist, y.first));
				}
			}
		}
	}
	return dist;
}
V<V<pin>> make_w(vvel v) {
	int n = v.size();
	V<V<pin>> ret(n);
	rep(i, n) {
		for (int x : v[i]) {
			ret[i].push_back(mkp(x, 1));
		}
	}
	return ret;
}
void make_tree(vvel& chi, vel& par, int n,int st) {
	V<V<pin>> way(n);
	rep(i, n - 1) {
		int a, b; cin >> a >> b; a--; b--;
		way[a].push_back(mkp(b, 1));
		way[b].push_back(mkp(a, 1));
	}
	vel dist = dijk(way, st, n + 1);
	par = vel(n, -1);
	chi = vvel(n);
	rep(i, n) {
		for (auto nex : way[i]) {
			int pot = nex.first;
			if (dist[pot] > dist[i]) { chi[i].push_back(pot); }
			else { par[i] = pot; }
		}
	}
}
void pri(vel& v) {
	if (v.size() == 0) { return; }
	cout << v[0];
	rep(i, v.size() - 1) { cout << " " << v[i + 1]; }
	cout << endl;
	return;
}
vvel disj_min(vel& v) {
	int n = v.size();
	vvel ret(22, vel(n));
	ret[0] = v;
	rep(i, 21) {
		rep(j, n) {
			int nex = j + (1 << i);
			if (nex < n) {
				ret[i + 1][j] = min(ret[i][j], ret[i][nex]);
			}
			else {
				ret[i + 1][j] = ret[i][j];
			}
		}
	}
	return ret;
}
int find_min(vvel& dv, int l, int r) {
	int i = 21;
	while (l + (1 << i) > r) {
		i--;
	}
	while (i >= 0) {
		if (dv[i][l] > dv[i][r - (1 << i)]) {
			l = r - (1 << i);
		}
		else {
			r = l + (1 << i);
		}
		i--;
	}
	return l;
}
V<V<pin>> dbl(V<pin>& v) {
	V<V<pin>> ans(20, V<pin>(v));
	int n = v.size();
	rep(i, 19) {
		rep(j, n) {
			ans[i + 1][j].first = ans[i][ans[i][j].first].first;
			ans[i + 1][j].second = max(ans[i][j].second, ans[i][ans[i][j].first].second);
		}
	}
	return ans;
}
int lca(int s, int t, int diff, V<V<pin>>& pa) {
	if (diff < 0) { return lca(t, s, -diff, pa); }
	int ans = 0;
	rep(i, 19) {
		if ((diff & (1 << i)) != 0) {
			mmax(ans, pa[i][s].second);
			s = pa[i][s].first;
		}
	}
	for (int i = 19; i >= 0; i--) {
		if (pa[i][s] != pa[i][t]) {
			mmax(ans, pa[i][s].second);
			s = pa[i][s].first;
			mmax(ans, pa[i][t].second);
			t = pa[i][t].first;
		}
	}
	if (s != t) {
		mmax(ans, pa[0][s].second);
		mmax(ans, pa[0][t].second);
	}
	return ans;
}
void alp(int n, vel& pr) {
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0) {
			pr.push_back(i);
			while (n % i == 0) { n /= i; }
		}
	}
	if (n != 1) { pr.push_back(n); }
}
vel dx = { 0,0,-1,1,1,-1 };
vel dy = { 1,-1,0,0,1,1 };
#define all(a) a.begin(),a.end()
template<typename T>
void mk_uni(V<T>& a) {
	std::sort(a.begin(), a.end());
	a.erase(std::unique(a.begin(), a.end()), a.end());
}
template <std::uint_fast64_t Modulus> class modint {
	using u64 = std::uint_fast64_t;

public:
	u64 a;

	constexpr modint(const u64 x = 0) noexcept : a(x% Modulus) {}
	constexpr u64& value() noexcept { return a; }
	constexpr const u64& value() const noexcept { return a; }
	constexpr modint operator+(const modint rhs) const noexcept {
		return modint(*this) += rhs;
	}
	constexpr modint operator-(const modint rhs) const noexcept {
		return modint(*this) -= rhs;
	}
	constexpr modint operator*(const modint rhs) const noexcept {
		return modint(*this) *= rhs;
	}
	constexpr modint operator/(const modint rhs) const noexcept {
		return modint(*this) /= rhs;
	}
	constexpr modint& operator+=(const modint rhs) noexcept {
		a += rhs.a;
		if (a >= Modulus) {
			a -= Modulus;
		}
		return *this;
	}
	constexpr modint& operator-=(const modint rhs) noexcept {
		if (a < rhs.a) {
			a += Modulus;
		}
		a -= rhs.a;
		return *this;
	}
	constexpr modint& operator*=(const modint rhs) noexcept {
		a = a * rhs.a % Modulus;
		return *this;
	}
	constexpr modint& operator/=(modint rhs) noexcept {
		u64 exp = Modulus - 2;
		while (exp) {
			if (exp % 2) {
				*this *= rhs;
			}
			rhs *= rhs;
			exp /= 2;
		}
		return *this;
	}
};
#pragma endregion
//using mint = modint<998244353>;
using mint = modint<1000000007>;
using vem = vector<mint>;
using vvem = vector<vem>;
#pragma region mint
mint mpow(mint a, int n) {
	mint ans = 1;
	while (n) {
		if (n & 1) { ans *= a; }
		a *= a; n /= 2;
	}
	return ans;
}
vem kai, inv_kai;
void make_kai(int n) {
	kai = vem(n + 1, 1);
	inv_kai = vem(n + 1, 1);
	rep(i, n) { kai[i + 1] = kai[i] * (i + 1); }
	inv_kai[n] = (mint)1 / kai[n];
	for (int i = n; i > 0; i--) { inv_kai[i - 1] = inv_kai[i] * i; }
}
mint com(int n, int r) {
	if (n < 0 || r < 0 || n < r) { return 0; }
	return kai[n] * inv_kai[r] * inv_kai[n - r];
}
mint per(int n, int r) {
	if (n < 0 || r < 0 || n < r) { return 0; }
	return kai[n] * inv_kai[r];
}
#pragma endregion
vel ans;
int inf = 20001*(long long)1000000;
signed main() {
	omajinai;
	int n; cin >> n;
	vel x(n), y(n),p(n);
	V<pin> qx(n), qy(n);
	rep(i, n) { 
		cin >> x[i] >> y[i]>>p[i]; 
		qx[i] = mkp(x[i], i);
		qy[i] = mkp(y[i], i);
	}
	sor(qx), sor(qy);
	ans=vel(n + 1, (n + 1)* inf);
	vel use(n, 0);
	int cnt = 0;
	int k = 1;
	rep(i, n) { k *= 3; } 
	rep(i, k) {
		vel use(n, 0);
		int mi = i;
		rep(j, n) { use[j] = mi % 3; mi /= 3; }
		int cnt = 0;
		vel ax = { -inf }; vel ay = ax;
		int mnx = 0; int mny = 0;
		bool flx = false, fly = false;
		rep(j, n) {
			int jx = qx[j].second;
			int jy = qy[j].second;
			if (!flx && x[jx] > 0) { ax.push_back(0); flx = true; }
			if (use[jx] == 1) { ax.push_back(x[jx]); cnt++; }
			if (!fly && y[jx] > 0) { ay.push_back(0); fly = true; }
			if(use[jy]==2){ ay.push_back(y[jy]); cnt++; }
		}
		if (!flx) { ax.push_back(0); }
		if (!fly) { ay.push_back(0); }
		ax.push_back(inf); ay.push_back(inf);
		auto itx = ax.begin(); auto itxp = itx; itxp++;
		auto ity = ay.begin(); auto ityp = ity; ityp++;
		vel mn(n, inf);
		rep(i, n) {
			int jx = qx[i].second; int vx = qx[i].first;
			int jy = qy[i].second; int vy = qy[i].first;
			while ((*itxp) <= vx) { itx++; itxp++; }
			mmin(mn[jx], (*itxp) - vx);
			mmin(mn[jx], vx - (*itx));
			while ((*ityp) <= vy) { ity++; ityp++; }
			mmin(mn[jy], (*ityp) -vy);
			mmin(mn[jy], vy - (*ity));
		}
		int res = 0;
		rep(i, n) { res += mn[i] * p[i]; }
		mmin(ans[cnt], res);
	}
	rep(i, n + 1) { cout << ans[i] << endl; }
	return 0;
}