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)

/*!enjoy karo yaar!*/

//CF,CC,AtC,SPOJ: hp1999
//HE: hemant269
//HR: hemant2132

#include<bits/stdc++.h>
using namespace std;

/*-----<Defines>-----*/
#define     int             long long int
#define     fast()          ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define     all(x)          begin(x),end(x)
#define     rz(x)           resize(x)
#define     asn(x,y)        assign(x,y)
#define     mem(a,b)        memset(a,b,sizeof(a))
#define     sz(x)           ((int)size(x))
#define     eb              emplace_back
#define     pb              push_back
#define     pf              push_front
#define     pob             pop_back
#define     pof             pop_front
#define     ins             insert
#define     vi              vector<int>
#define     pii             pair<int,int>
#define     mii             map<int,int>
#define     F               first
#define     S               second
#define     remax(a,b)      a=max(a,b)
#define     remin(a,b)      a=min(a,b)
#define     bitcount(x)     __builtin_popcountll(x)
#define     iceil(n,x)      (((n)-1)/(x)+1)
#define     flush           fflush(stdout)

using ull=unsigned long long;
using ll=long long;
using ld=long double;
/*-----</Defines>-----*/

/*-----<Debugging Stuff>-----*/
string to_string(char s)
{   string res="'''";res[1]=s;return res; }

string to_string(string s)
{   return '"'+s+'"'; }

string to_string(const char* s)
{   return to_string((string)s); }

string to_string(bool b)
{   return (b?"true":"false"); }

template<typename A, typename B>
string to_string(pair<A, B> p)
{   return "("+to_string(p.F)+", "+to_string(p.S)+")"; }

template<typename A>
string to_string(A v)
{
    bool first=1; string res="{";
    for(const auto &x:v)
    {
        if(!first)  res+=", ";
        first=0;
        res+=to_string(x);
    }
    res+="}";
    return res;
}

void debug() { cout<<"\n"; }
template<typename Head, typename... Tail>
void debug(Head H, Tail... T)
{
    cout<<" "<<to_string(H);
    debug(T...);
}

#define show(...) cout << "[" << #__VA_ARGS__ << "]:", debug(__VA_ARGS__)
/*-----</Debugging Stuff>-----*/

/*-----<Mod Operations>-----*/
const int M=1e9+7;
inline int add(int a,int b,int p=M){ a=a+b; if(a>=p) a-=p; return a; }
inline int sub(int a,int b,int p=M){ a=a-b; if(a<0) a+=p; return a; }
inline int mul(int a,int b,int p=M){ a=(a*1ll*b)%p; return a; }
/*-----</Mod Operations>-----*/

/*-----<Constants>-----*/
const ld pi=acos(-1),eps=1e-9;
const ll inf=1e18;
const int N=1;
/*-----</Constants>-----*/

int pw(int x,int n)
{
    int res=1;

    for(;n;n/=2)
    {
        if(n%2)
            res=(res*x);

        x=(x*x);
    }

    return res;
}

void solve()
{
    int n;
    cin>>n;

    vector<pii> v(n);
    vi popu(n);
    for(int i=0;i<n;++i)
    {
        cin>>v[i].F>>v[i].S;
        cin>>popu[i];
    }

    int tot=pw(3,n)-1;

//    show(tot);

    vi ans(n+1,inf);
    vi tmp,px,py;
    for(int i=0;i<=tot;++i)
    {
        px.clear();py.clear();
        px.pb(0);py.pb(0);

        int cnt=0;
        for(int z=i,j=0;z;z/=3,++j)
        {
            if(z%3)
            {
                ++cnt;
                if(z%3==1) px.pb(v[j].F);
                else py.pb(v[j].S);
            }
        }

        if(!ans[cnt])
            continue;

        sort(all(px));
        sort(all(py));

        int cur=0;
        int pos,dist;
        for(int j=0;j<n;++j)
        {
            auto [x,y]=v[j];

            dist=inf;

            pos=upper_bound(all(px),x)-begin(px);
            if(pos<sz(px)) remin(dist,abs(x-px[pos]));
            if(pos-1>=0) remin(dist,abs(x-px[pos-1]));

            pos=upper_bound(all(py),y)-begin(py);
            if(pos<sz(py)) remin(dist,abs(y-py[pos]));
            if(pos-1>=0) remin(dist,abs(y-py[pos-1]));

            cur+=dist*popu[j];
        }

        remin(ans[cnt],cur);
    }

    ans[n]=0;
    for(auto c:ans)
        cout<<c<<"\n";
}

int32_t main()
{
    fast();

    int t=1;
    //cin>>t;
    for(int z=1;z<=t;++z)
    {
        solve();
        //cout<<"\n";
    }

    return 0;
}