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 <bits/stdc++.h>
using namespace std;
#if __has_include(<boost/multiprecision/cpp_int.hpp>)
	#include <boost/multiprecision/cpp_int.hpp>
	#include <boost/multiprecision/cpp_dec_float.hpp>
	using bll = boost::multiprecision::cpp_int;
	using bdouble = boost::multiprecision::number<boost::multiprecision::cpp_dec_float<100>>;
	using namespace boost::multiprecision;
#endif
#if __has_include(<boost/stacktrace.hpp>)
	#define BOOST_STACKTRACE_USE_ADDR2LINE
	#define BOOST_STACKTRACE_ADDR2LINE_LOCATION /usr/local/opt/binutils/bin/addr2line
	#define _GNU_SOURCE 1
	#include <boost/stacktrace.hpp>
#endif
#ifdef LOCAL_TEST
	namespace std {
		template<typename T> class dvector : public std::vector<T> {
		public:
			dvector() : std::vector<T>() {}
			explicit dvector(size_t n, const T& value = T()) : std::vector<T>(n, value) {}
			dvector(const std::vector<T>& v) : std::vector<T>(v) {}
			dvector(const std::initializer_list<T> il) : std::vector<T>(il) {}
			dvector(const std::string::iterator first, const std::string::iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::iterator first, const typename std::vector<T>::iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::reverse_iterator first, const typename std::vector<T>::reverse_iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::const_iterator first, const typename std::vector<T>::const_iterator last) : std::vector<T>(first, last) {}
			dvector(const typename std::vector<T>::const_reverse_iterator first, const typename std::vector<T>::const_reverse_iterator last) : std::vector<T>(first, last) {}
			T& operator[](size_t n) {
				try { return this->at(n); } catch (const std::exception& e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; return this->at(n); }
			}
			const T& operator[](size_t n) const {
				try { return this->at(n); } catch (const std::exception& e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; return this->at(n); }
			}
		};
	}
	class dbool {
	private:
		bool boolvalue;
	public:
		dbool() : boolvalue(false) {}
		dbool(bool b) : boolvalue(b) {}
		operator bool&() { return boolvalue; }
		operator const bool&() const { return boolvalue; }
	};
	#define vector dvector
	#define bool dbool
#endif
#ifdef LOCAL_DEV
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::vector<T>& v) {
		for (size_t i = 0; i < v.size(); ++i){ s << v[i]; if (i < v.size() - 1) s << "\t"; } return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::vector<std::vector<T>>& vv) {
		s << "\\\n"; for (size_t i = 0; i < vv.size(); ++i){ s << vv[i] << "\n"; } return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::deque<T>& v) {
		for (size_t i = 0; i < v.size(); ++i){ s << v[i]; if (i < v.size() - 1) s << "\t"; } return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::set<T>& se) {
		s << "{ "; for (auto itr = se.begin(); itr != se.end(); ++itr){ s << (*itr) << "\t"; } s << "}"; return s; }
	template<typename T> std::ostream& operator<<(std::ostream& s, const std::multiset<T>& se) {
		s << "{ "; for (auto itr = se.begin(); itr != se.end(); ++itr){ s << (*itr) << "\t"; } s << "}"; return s; }
	template <typename T, size_t N> std::ostream& operator<<(std::ostream& s, const std::array<T, N>& a) {
		s << "{ "; for (size_t i = 0; i < N; ++i){ s << a[i] << "\t"; } s << "}"; return s; }
	template<typename T1, typename T2> std::ostream& operator<<(std::ostream& s, const std::map<T1, T2>& m) {
		s << "{\n"; for (auto itr = m.begin(); itr != m.end(); ++itr){ s << "\t" << (*itr).first << " : " << (*itr).second << "\n"; } s << "}"; return s; }
	template<typename T1, typename T2> std::ostream& operator<<(std::ostream& s, const std::pair<T1, T2>& p) {
		return s << "(" << p.first << ", " << p.second << ")"; }
	class SIGFPE_exception : std::exception {};
	class SIGSEGV_exception : std::exception {};
	void catch_SIGFPE([[maybe_unused]] int e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; throw SIGFPE_exception(); }
	void catch_SIGSEGV([[maybe_unused]] int e) { std::cerr << boost::stacktrace::stacktrace() << '\n'; throw SIGSEGV_exception(); }
	signed convertedmain();
	signed main() { signal(SIGFPE, catch_SIGFPE); signal(SIGSEGV, catch_SIGSEGV); return convertedmain(); }
	#define main() convertedmain()
	void debug_impl() { std::cerr << '\n'; }
	template<typename Head, typename... Tail> void debug_impl(Head head, Tail... tail) { std::cerr << " " << head << (sizeof...(tail) ? "," : ""); debug_impl(tail...); }
	#define debug(...) do { std::cerr << "(" << #__VA_ARGS__ << ") ="; debug_impl(__VA_ARGS__); } while (false)
#else
	#define debug(...) do {} while (false)
#endif
//#define int long long
using ll = long long;
//constexpr int INF = (ll)1e9 + 7;//INT_MAX=(1<<31)-1=2147483647
constexpr ll INF = (ll)1e18;//(1LL<<63)-1=9223372036854775807
constexpr ll MOD = (ll)1e9 + 7;
constexpr double EPS = 1e-9;
constexpr ll dx[4] = {1, 0, -1, 0};
constexpr ll dy[4] = {0, 1, 0, -1};
constexpr ll dx8[8] = {1, 0, -1, 0, 1, 1, -1, -1};
constexpr ll dy8[8] = {0, 1, 0, -1, 1, -1, 1, -1};
#define rep(i, n)   for(ll i=0, i##_length=(n); i< i##_length; ++i)
#define repeq(i, n) for(ll i=1, i##_length=(n); i<=i##_length; ++i)
#define rrep(i, n)   for(ll i=(n)-1; i>=0; --i)
#define rrepeq(i, n) for(ll i=(n)  ; i>=1; --i)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
void p() { std::cout << '\n'; }
template<typename Head, typename... Tail> void p(Head head, Tail... tail) { std::cout << head << (sizeof...(tail) ? " " : ""); p(tail...); }
template<typename T> inline void pv(std::vector<T>& v) { for(ll i=0, N=v.size(); i<N; i++) std::cout << v[i] << " \n"[i==N-1]; }
template<typename T> inline bool chmax(T& a, T b) { return a < b && (a = b, true); }
template<typename T> inline bool chmin(T& a, T b) { return a > b && (a = b, true); }
template<typename T> inline void uniq(std::vector<T>& v) { v.erase(std::unique(v.begin(), v.end()), v.end()); }

/*-----8<-----template-----8<-----*/

// (0..n-1)のn個の要素について、nCkの組み合わせに対して関数fを適用する
template<typename Function> void foreach_comb(ll n, ll k, Function f) {
	vector<ll> indexes(k);
	auto recursive_comb = [&](auto&& recursive_comb, vector<ll> &indexes, ll s, ll rest, Function& f){
		if (rest == 0) {
			f(indexes);
		} else {
			if (s < 0) return;
			recursive_comb(recursive_comb, indexes, s - 1, rest, f);
			indexes[rest - 1] = s;
			recursive_comb(recursive_comb, indexes, s - 1, rest - 1, f);
		}
	};
	recursive_comb(recursive_comb, indexes, n - 1, k, f);
}

/*-----8<-----library-----8<-----*/

constexpr ll N_MAX = 15;
constexpr ll X_I_MAX = 10000;
constexpr ll Y_I_MAX = 10000;
constexpr ll P_I_MAX = 1000000;

ll N;
vector<ll> X;
vector<ll> Y;
vector<ll> P;

void solve() {
	cin >> N;
	X = vector<ll>(N);
	Y = vector<ll>(N);
	P = vector<ll>(N);
	rep(i, N) { cin >> X[i] >> Y[i] >> P[i]; }
	vector<ll> anslist(N+1,INF);

	//bit全探索 K^N通りループ
	ll bitsize = N;
	vector<ll> bitlen(bitsize+1, 3);
	vector<ll> bit(bitsize+1, 0);
	auto f=[&](auto&& f,vector<ll> &bit)->void{
		if(bit.size()==(size_t)N){
			ll honsu=0;
			rep(i,bitsize)if(bit[i]<=1)honsu++;

			set<ll> tate,yoko;
			tate.insert(0);
			yoko.insert(0);
			rep(i, bitsize) {
				if (bit[i]==0) {
					tate.insert(X[i]);
				}else if (bit[i]==1) {
					yoko.insert(Y[i]);
				}
			}
			ll tmp=0;
			rep(i,N){
				auto it1=tate.lower_bound(X[i]);
				ll t=INF;
				if(it1!=tate.end()){
					//debug(*it1);
					chmin(t,abs(X[i]-*it1));
				}
				if(it1!=tate.begin()){
					advance(it1,-1);
					chmin(t,abs(X[i]-*it1));
				}
				auto it2=yoko.lower_bound(Y[i]);
				ll s=INF;
				if(it2!=yoko.end()){
					//debug(*it2);
					chmin(s,abs(Y[i]-*it2));
				}
				if(it2!=yoko.begin()){
					advance(it2,-1);
					chmin(s,abs(Y[i]-*it2));
				}
				tmp+=min(t,s)*P[i];
			}
			debug(honsu,bit,tmp);
			chmin(anslist[honsu],tmp);

			//for(ll i=0; ++bit[i]==bitlen[i]; i++)bit[i]=0;
		}else{
			rep(i,3){
				bit.push_back(i);
				f(f,bit);
				bit.pop_back();
			}
		}
	};
	vector<ll> x;
	f(f,x);

	rep(i,N+1)cout << anslist[i] << endl;
}

// https://atcoder.jp/contests/m-solutions2020/tasks/m_solutions2020_e
signed main() {
	solve();
	return 0;
}