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)

#include <iostream>
#include <string>
#include <math.h>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <algorithm>

using namespace std;

using ll = long long;
#define int ll

// [a -> b-1]
#define reps(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
// [0 -> a-1]
#define rep(i, a) reps(i, 0, (a))
// [a-1 -> b]
#define rreps(i, a, b) for (int i = (int)((a)-1); i >= (int)(b); i--)
// [a-1 -> 0]
#define rrep(i, a) rreps(i, a, 0)
#define all(v) (v).begin(), (v).end()
// next_permutation(all(v))
#define PERM(v) next_permutation(all(v))
/*sort(all(v));
 * (v).erase(unique(all(v)), v.end())*/
#define UNIQUE(v)\
	sort(all(v));\
	(v).erase(unique(all(v)), v.end())
// b is [a, c)
#define RANGE(a, b, c) ((a) <= (b) && (b) < (c))
// c is [a, e) && d is [b, f)
#define RANGE2D(a, b, c, d, e, f) (RANGE((a), (c), (e)) && RANGE((b), (d), (f)))
#define chmin(a, b) a = min(a, (b))
#define chmax(a, b) a = max(a, (b))
#define fcout cout << fixed << setprecision(15)
#define YES(f) cout << ((f) ? YES_STR : NO_STR) << endl;

#ifdef LOCAL_DEBUG_OPTION
namespace debug_print_n{
struct DebugPrint {
	template<class T, class U = remove_const_t<remove_reference_t<T>>>
	static enable_if_t<is_integral<U>::value && !is_same<U, bool>::value && !is_same<U, char>::value> DUMP_SUB_FUNCTION(T x) {
		cerr << x;
	}
	template<class T, class U = remove_const_t<remove_reference_t<T>>>
	static enable_if_t<is_same<U, bool>::value> DUMP_SUB_FUNCTION(T x) {
		if (x) cerr << "True";
		else cerr << "False";
	}
	template<class T, class U = remove_const_t<remove_reference_t<T>>>
	static enable_if_t<is_same<U, char>::value> DUMP_SUB_FUNCTION(T x) {
		cerr << '\'' << x << '\'';
	}
	template<class T, class U = remove_const_t<remove_reference_t<T>>>
	static enable_if_t<is_same<U, string>::value> DUMP_SUB_FUNCTION(T x) {
		cerr << '\"' << x << '\"';
	}
	template<class T>
	static enable_if_t<is_same<T, const char*>::value> DUMP_SUB_FUNCTION(T x) {
		cerr << string(x);
	}
	template<class T>
	static enable_if_t<true> DUMP_SUB_FUNCTION(vector<T> x) {
		cerr << '{';
		for (size_t i = 0; i < x.size(); i++) {
			if (i) cerr << ", ";
			DUMP_SUB_FUNCTION(x[i]);
		}
		cerr << '}';
	}
	template<class T, size_t N>
	static enable_if_t<true> DUMP_SUB_FUNCTION(array<T, N> x) {
		cerr << '{';
		for (size_t i = 0; i < x.size(); i++) {
			if (i) cerr << ", ";
			DUMP_SUB_FUNCTION(x[i]);
		}
		cerr << '}';
	}
	template<class T>
	static enable_if_t<true> DUMP_SUB_FUNCTION(set<T> x) {
		cerr << '[';
		size_t i = 0;
		for (auto itr = x.begin(); itr != x.end(); itr++, i++) {
			if (i) cerr << ", ";
			DUMP_SUB_FUNCTION(*itr);
		}
		cerr << ']';
	}
	template<class T, class U>
	static enable_if_t<true> DUMP_SUB_FUNCTION(map<T, U> x) {
		cerr << '[';
		size_t i = 0;
		for (auto itr = x.begin(); itr != x.end(); itr++, i++) {
			if (i) cerr << ", ";
			DUMP_SUB_FUNCTION(itr->first);
			cerr << " : ";
			DUMP_SUB_FUNCTION(itr->second);
		}
		cerr << ']';
	}
	template<class T, class U = remove_const_t<remove_reference_t<T>>>
	static enable_if_t<is_floating_point<U>::value> DUMP_SUB_FUNCTION(T x) {
		cerr << fixed << setprecision(5) << x;
	}
	template<class T, class U>
	static enable_if_t<true> DUMP_SUB_FUNCTION(pair<T, U> p) {
		cerr << '(';
		DUMP_SUB_FUNCTION(p.first);
		cerr << ", ";
		DUMP_SUB_FUNCTION(p.second);
		cerr << ')';
	}
	template<int num = 0, class... Args>
	static enable_if_t<num == sizeof...(Args)> DUMP_SUB_FUNCTION(tuple<Args...> t) {
		if (num) cerr << ')';
	}
	template<int num = 0, class... Args>
	static enable_if_t<num < sizeof...(Args)> DUMP_SUB_FUNCTION(tuple<Args...> t) {
		if (num == 0) cerr << '(';
		DUMP_SUB_FUNCTION(get<num>(t));
		if (num + 1 < sizeof...(Args)) cerr << ", ";
		DUMP_SUB_FUNCTION<num+1>(t);
	}
	template<long unsigned N>
	static enable_if_t<true> DUMP_SUB_FUNCTION(bitset<N> x) {
		cerr << x;
	}
};
void dump() {
	cerr << endl;
}
template<class Head, class... Tail>
void dump(Head&& head, Tail&&... tail) {
	DebugPrint::DUMP_SUB_FUNCTION(head);
	if (sizeof...(Tail) != 0) {
		cerr << ' ';
	}
	dump(forward<Tail>(tail)...);
}
} // namespace debug_print_n
using debug_print_n::dump;
#else
template<class... Args> void dump(Args&&... args) { }
#endif

namespace bit {
using u32 = uint_fast32_t;
using u64 = unsigned long long;
template<class T>
constexpr T set(T bin, u32 pos) {
	return bin | (1ull << pos);
}
template<class T>
constexpr T unset(T bin, u32 pos) {
	return bin & ~(1ull << pos);
}
template<class T>
constexpr T get(T bin, u32 pos) {
	return bin & (1ull << pos);
}
template<class T>
constexpr bool check(T bin, u32 pos) {
	return get(bin, pos);
}
constexpr u32 popcount(u64 bin) {
	return __builtin_popcountll(bin);
}
} // namespace bit


signed main(){
	int N; cin >> N;
	int N3 = 1;
	rep(i, N) N3 *= 3;
	vector<int> res(N+1, 4e18);
	vector<int> X(N), Y(N), P(N);
	rep(i, N) cin >> X[i] >> Y[i] >> P[i];
	rep(i, N3) {
		vector<int> R{(int)-1e9, 0, (int)1e9}, C{(int) -1e9, 0, (int)1e9};
		vector<int> l;
		int t = i;
		int cnt = N;
		rep(j, N) {
			if (t % 3 == 0) R.push_back(Y[j]);
			else if (t % 3 == 1) C.push_back(X[j]);
			else cnt--, l.push_back(j);
			t /= 3;
		}
		sort(all(R));
		sort(all(C));
		int r = 0;
		for (auto j : l) {
			auto f = [&](auto itr, int x) {
				return abs(*itr - x) * P[j];
			};
			auto a = upper_bound(all(R), Y[j]);
			auto b = prev(a);
			auto c = upper_bound(all(C), X[j]);
			auto d = prev(c);
			r += min({f(a, Y[j]), f(b, Y[j]), f(c, X[j]), f(d, X[j])});
		}
		chmin(res[cnt], r);
	}
	rep(i, N+1) {
		cout << res[i] << endl;
	}
	return 0;
}