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;
#define ll long long
#define all(V) V.begin(),V.end()
const ll MOD = 1000000007;

vector<ll> X, Y, P;
vector<pair<ll, ll>> cx, cy;
int N;
bool xg[18], yg[18];
ll D[18];
ll R[18];

void chmin(ll& A, ll B) {
	A = min(A, B);
}

void dfs(int c, int k) {
	if (c == N) {
		for (int i = 0;i < N;i++)D[i] = min(abs(X[i]), abs(Y[i]));
		ll mv = -MOD;
		for (int i = 0;i < N;i++) {
			if (xg[cx[i].second]) {
				mv = cx[i].first;
			}
			chmin(D[cx[i].second], cx[i].first - mv);
		}
		mv = MOD;
		for (int i = N - 1;i >= 0;i--) {
			if (xg[cx[i].second]) {
				mv = cx[i].first;
			}
			chmin(D[cx[i].second], mv - cx[i].first);
		}
		mv = -MOD;
		for (int i = 0;i < N;i++) {
			if (yg[cy[i].second]) {
				mv = cy[i].first;
			}
			chmin(D[cy[i].second], cy[i].first - mv);
		}
		mv = MOD;
		for (int i = N - 1;i >= 0;i--) {
			if (yg[cy[i].second]) {
				mv = cy[i].first;
			}
			chmin(D[cy[i].second], mv - cy[i].first);
		}

		ll S = 0;
		for (int i = 0;i < N;i++) {
			S += D[i] * P[i];
		}
		chmin(R[k], S);
		return;
	}

	xg[c] = 1;
	dfs(c + 1, k + 1);
	xg[c] = 0;

	yg[c] = 1;
	dfs(c + 1, k + 1);
	yg[c] = 0;

	dfs(c + 1, k);
}

int main() {
	cin >> N;
	X = Y = P = vector<ll>(N);
	cx = cy = vector<pair<ll, ll>>(N);

	for (int i = 0;i < N;i++) {
		cin >> X[i] >> Y[i] >> P[i];
		cx[i] = { X[i],i };
		cy[i] = { Y[i],i };
		xg[i] = yg[i] = 0;
		R[i] = INT64_MAX;
	}
	sort(all(cx));
	sort(all(cy));

	R[N] = INT64_MAX;
	dfs(0, 0);

	for (int i = 0;i <= N;i++) {
		cout << R[i] << endl;
	}
}