#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define pb push_back
#define all(in) in.begin(),in.end()

template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }

using ll = long long;
using namespace std;
const ll INF = 100100100100100;

int main(){
    // 入力
    ll n; cin >> n;
    vector<ll> x(n), y(n), p(n), ans(n+1,INF);
    rep(i,n) cin >> x[i] >> y[i] >> p[i];
    // 前処理
    vector<vector<ll>> prex(1<<n,vector<ll>(n)), prey(1<<n,vector<ll>(n));
    rep(i,1<<n){
        rep(j,n) { prex[i][j] = abs(x[j]); prey[i][j] = abs(y[j]); }
        rep(j,n) if(i & 1<<j) rep(k,n) { chmin(prex[i][k],abs(x[k]-x[j])); chmin(prey[i][k],abs(y[k]-y[j])); }
    }
    // 本処理
    ll count = 1; rep(i,n) count *= 3;
    rep(i,count){
        ll xnum = 0, ynum = 0;
        ll num = 1, bit = n, save = i, save2;
        rep(j,n){
          save2 = save % 3; save /= 3;
            if(save2 == 0) bit--;
            else if(save2 == 1) xnum += num;
            else if(save2 == 2) ynum += num;
            num *= 2;
        }
        ll change = 0; rep(j,n) change += min(prex[xnum][j],prey[ynum][j]) * p[j];
        chmin(ans[bit],change);
    }
    rep(i,n+1) cout << ans[i] << endl;
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define pb push_back
#define all(in) in.begin(),in.end()

template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }

using ll = long long;
using namespace std;
const ll INF = 100100100100100;

int main(){
    // 入力
    ll n; cin >> n;
    vector<ll> x(n), y(n), p(n), ans(n+1,INF);
    rep(i,n) cin >> x[i] >> y[i] >> p[i];
    // 前処理
    vector<vector<ll>> prex(1<<n,vector<ll>(n)), prey(1<<n,vector<ll>(n));
    rep(i,1<<n){
        rep(j,n) { prex[i][j] = abs(x[j]); prey[i][j] = abs(y[j]); }
        rep(j,n) if(i & 1<<j) rep(k,n) { chmin(prex[i][k],abs(x[k]-x[j])); chmin(prey[i][k],abs(y[k]-y[j])); }
    }
    // 本処理
    ll count = 1; rep(i,n) count *= 3;
    rep(i,count){
        ll xnum = 0, ynum = 0;
        ll num = 1, bit = n, save = i, save2;
        rep(j,n){
          save2 = save % 3; save /= 3;
            if(save2 == 0) bit--;
            else if(save2 == 1) xnum += num;
            else if(save2 == 2) ynum += num;
            num *= 2;
        }
        ll change = 0; rep(j,n) change += min(prex[xnum][j],prey[ynum][j]) * p[j];
        chmin(ans[bit],change);
    }
    rep(i,n+1) cout << ans[i] << endl;
    return 0;
}
```