#include <bits/stdc++.h>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

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 (a>b) { a=b; return 1; } return 0; }

int n;
vector<ll> x, y, p;
vector<ll> ans;

vector<vector<ll>> cx, cy;

void dfs(int k, int d, vector<int> &w){
	if(w.size() == n){
		ll xb = 0, yb = 0;
		rep(i, n){
			if(w[i] == 1) xb |= (1 << i);
			if(w[i] == 2) yb |= (1 << i);
		}

		ll t = 0;
		rep(i, n){
			t += p[i] * min(cx[i][xb], cy[i][yb]);
		}
		chmin(ans[d], t);
		return;
	}

	w.push_back(0);
	dfs(k, d, w);
	w.pop_back();

	w.push_back(1);
	dfs(k, d+1, w);
	w.pop_back();

	w.push_back(2);
	dfs(k, d+1, w);
	w.pop_back();
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n;

	x.resize(n);
	y.resize(n);
	p.resize(n);
	rep(i, n) cin >> x[i] >> y[i] >> p[i];

	cx.resize(n, vector<ll>((1 << n), 0));
	cy.resize(n, vector<ll>((1 << n), 0));
	
	rep(i, n){
		for(ll bit = 0; bit < (1 << n); bit++){
			ll t = abs(x[i] - 0);
			rep(j, n){
				if((bit >> j) & 1) chmin(t, abs(x[i] - x[j]));
			}
			cx[i][bit] = t;
		}
	}

	rep(i, n){
		for(ll bit = 0; bit < (1 << n); bit++){
			ll t = abs(y[i] - 0);
			rep(j, n){
				if((bit >> j) & 1) chmin(t, abs(y[i] - y[j]));
			}
			cy[i][bit] = t;
		}
	}

	ans.resize(n+1, 1e18);
	vector<int> w;
	dfs(0, 0, w);
	rep(i, n+1) cout << ans[i] << endl;
}

### Prompt

Please formulate a CPP solution to 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>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

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 (a>b) { a=b; return 1; } return 0; }

int n;
vector<ll> x, y, p;
vector<ll> ans;

vector<vector<ll>> cx, cy;

void dfs(int k, int d, vector<int> &w){
	if(w.size() == n){
		ll xb = 0, yb = 0;
		rep(i, n){
			if(w[i] == 1) xb |= (1 << i);
			if(w[i] == 2) yb |= (1 << i);
		}

		ll t = 0;
		rep(i, n){
			t += p[i] * min(cx[i][xb], cy[i][yb]);
		}
		chmin(ans[d], t);
		return;
	}

	w.push_back(0);
	dfs(k, d, w);
	w.pop_back();

	w.push_back(1);
	dfs(k, d+1, w);
	w.pop_back();

	w.push_back(2);
	dfs(k, d+1, w);
	w.pop_back();
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> n;

	x.resize(n);
	y.resize(n);
	p.resize(n);
	rep(i, n) cin >> x[i] >> y[i] >> p[i];

	cx.resize(n, vector<ll>((1 << n), 0));
	cy.resize(n, vector<ll>((1 << n), 0));
	
	rep(i, n){
		for(ll bit = 0; bit < (1 << n); bit++){
			ll t = abs(x[i] - 0);
			rep(j, n){
				if((bit >> j) & 1) chmin(t, abs(x[i] - x[j]));
			}
			cx[i][bit] = t;
		}
	}

	rep(i, n){
		for(ll bit = 0; bit < (1 << n); bit++){
			ll t = abs(y[i] - 0);
			rep(j, n){
				if((bit >> j) & 1) chmin(t, abs(y[i] - y[j]));
			}
			cy[i][bit] = t;
		}
	}

	ans.resize(n+1, 1e18);
	vector<int> w;
	dfs(0, 0, w);
	rep(i, n+1) cout << ans[i] << endl;
}

```