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

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
const long long INF = 1e18;
using ll = long long;
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define chmin(x,y) x = min(x,y);
long long dp[20][20];
using P = pair<int,int>;

vector<ll> calc(vector<P>& a) {
	int n = a.size();
	sort(a.begin(), a.end());
	rep(i, n + 1)rep(j, n + 1) dp[i][j] = INF;
	dp[0][0] = 0;
	rep(l, n) {
		for (int r = l + 1; r <= n; ++r) {
			{
				ll now = INF;
				for (int k = l; k < r; ++k) {
					ll s = 0;
					for (int i = l; i < r; ++i) {
						s += abs(a[i].first - a[k].first)*ll(a[i].second);
					}
					chmin(now, s);
				}
				rep(j, n) chmin(dp[r][j + 1], dp[l][j] + now);
			}
			{
				ll now = 0;
				for (int i = l; i < r; ++i) {
					now += abs(a[i].first)*ll(a[i].second);
				}
				rep(j, n + 1) chmin(dp[r][j], dp[l][j] + now);
			}
		}
	}
	vector<ll> res(n + 1, INF);
	rep(j, n + 1) res[j] = dp[n][j];
	return res;
}

int main() {
	int N;
	cin >> N;
	vector<int>X(N), Y(N), P(N);
	vector<long long >Ans(N + 1, INF);
	for (int n = 0;n<N; ++n) {
		cin >> X[n] >> Y[n] >> P[n];
	}
	for (int n = 0;n<(1<<N);++n) {
		vector<pair<int, int>>Ax,Ay;
		for (int m = 0;m<N;++m) {
			if (n>>m&1) {
				Ax.emplace_back(X[m],P[m]);
			}
			else {
				Ay.emplace_back(Y[m], P[m]);
			}
		}
		vector<long long> dx = calc(Ax); 
		vector<long long> dy = calc(Ay);
		for (int p = 0;p<dx.size();++p) {
			for (int q = 0;q<dy.size();++q) {
				Ans[p + q] = min(Ans[p + q], dx[p] + dy[q]);
			}
		}
	}
	for (int i = 0; i < N+1; ++i) {
		cout << Ans[i]<<endl;
	}
	return 0;
}