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<iostream>
#include<algorithm>
#include<cmath>

using namespace std;

using ll = long long;

int main() {
	int n;
	cin >> n;
	ll* x = new ll[n] ;
	ll* y = new ll[n];
	ll* p = new ll[n];
	ll* cost = new ll[n + 1];
	for (int i = 0;i < n;i++) {
		cin >> x[i] >> y[i] >> p[i];
		cost[i] = 1e18;
	}
	cost[n] = 1e18;
	long long** tate = new long long*[n];
	long long** yoko = new long long*[n];
	for (int i = 0;i < n;i++) {
		tate[i] = new long long[1 << n];
		yoko[i] = new long long[1 << n];
		for (int j = 0;j < 1 << n;j++) {
			tate[i][j] = abs(x[i]) * p[i];
			yoko[i][j] = abs(y[i]) * p[i];
		}
	}
	// calc tate and yoko
	for (int i = 0;i < n;i++) {
		for (int j = 0;j < 1 << n;j++) {
			for (int k = 0;k < n;k++) {
				if (j & 1 << k) {
					tate[i][j] = min(tate[i][j], abs(x[k] - x[i]) * p[i]);
					yoko[i][j] = min(yoko[i][j], abs(y[k] - y[i]) * p[i]);
				}
			}
		}
	}
	int* ind = new int[n];
	for (int i = 0;i < n;i++) {
		ind[i] = 0;
	}
	bool flag = true;;
	while (flag) {
		//syori
		int ind_tate, ind_yoko;
		ind_tate = ind_yoko = 0;
		int honsu = 0;
		for (int i = 0;i < n;i++) {
			if (ind[i] == 1) {
				ind_tate += 1<<i;
				honsu++;
			}
			if (ind[i] == 2) {
				ind_yoko += 1 << i;
				honsu++;
			}
		}
		ll c = 0;
		for (int i = 0;i < n;i++) {
			c += min(tate[i][ind_tate], yoko[i][ind_yoko]);
		}
		cost[honsu] = min(c, cost[honsu]);
		//ind update
		flag = false;
		for (int i = 0;i < n;i++) {
			if (ind[i] < 2) {
				ind[i]++;
				flag = true;
				break;
			}
			else {
				ind[i] = 0;
			}
		}

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