#include <bits/stdc++.h>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
#define ALL(a) a.begin(), a.end()
#define fi first
#define se second
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,Q,K,A,B;
string S;
//const ll MOD = 998244353;
const ll MOD = (1e+9) + 7;
typedef pair<ll, ll> P;
const ll INF = (1LL<<58);

template<class T> bool chmin(T &a, const T &b){
    if(a > b) {a = b; return true;}
    else return false;
}
template<class T> bool chmax(T &a, const T &b){
    if(a < b) {a = b; return true;}
    else return false;
}

vec x(15), y(15), p(15), ans(16, INF);

void dfs(vec &choose, vec &see, int now){
    ll sz = choose.size();
    if(now == N){
        ll res = 0;
        for(ll v : see){
            ll dist = min(abs(x[v]), abs(y[v]));
            for(ll to : choose){
                chmin(dist, to >= N ? abs(y[to - N] - y[v]) : abs(x[to] - x[v]));
            }
            res += dist * p[v];
        }
        chmin(ans[sz], res);
    }else{
        see.push_back(now);
        dfs(choose, see, now + 1);
        see.pop_back();
        choose.push_back(now);
        dfs(choose, see, now + 1);
        choose.pop_back();
        choose.push_back(now + N);
        dfs(choose, see, now + 1);
        choose.pop_back();

    }
}

int main() {
    cin>>N;
    rep(i,N) cin>>x[i]>>y[i]>>p[i];
    vec choose(0), see(0);
    dfs(choose, see, 0);
    rep(i,N + 1) cout<<ans[i]<<endl;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

Examples

Input

3
1 2 300
3 3 600
1 4 800


Output

2900
900
0
0


Input

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400


Output

13800
1600
0
0
0
0


Input

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200


Output

26700
13900
3200
1200
0
0
0


Input

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


Output

2576709
1569381
868031
605676
366338
141903
0
0
0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
#define ALL(a) a.begin(), a.end()
#define fi first
#define se second
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,Q,K,A,B;
string S;
//const ll MOD = 998244353;
const ll MOD = (1e+9) + 7;
typedef pair<ll, ll> P;
const ll INF = (1LL<<58);

template<class T> bool chmin(T &a, const T &b){
    if(a > b) {a = b; return true;}
    else return false;
}
template<class T> bool chmax(T &a, const T &b){
    if(a < b) {a = b; return true;}
    else return false;
}

vec x(15), y(15), p(15), ans(16, INF);

void dfs(vec &choose, vec &see, int now){
    ll sz = choose.size();
    if(now == N){
        ll res = 0;
        for(ll v : see){
            ll dist = min(abs(x[v]), abs(y[v]));
            for(ll to : choose){
                chmin(dist, to >= N ? abs(y[to - N] - y[v]) : abs(x[to] - x[v]));
            }
            res += dist * p[v];
        }
        chmin(ans[sz], res);
    }else{
        see.push_back(now);
        dfs(choose, see, now + 1);
        see.pop_back();
        choose.push_back(now);
        dfs(choose, see, now + 1);
        choose.pop_back();
        choose.push_back(now + N);
        dfs(choose, see, now + 1);
        choose.pop_back();

    }
}

int main() {
    cin>>N;
    rep(i,N) cin>>x[i]>>y[i]>>p[i];
    vec choose(0), see(0);
    dfs(choose, see, 0);
    rep(i,N + 1) cout<<ans[i]<<endl;
}

```