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;

int n, x[20], y[20];
int m[20];
long long rlt[20], w[20];

int v[20], szv, vv[20], szvv;

void DFS(int c){
	if(c == n){
		int cn = 0;

		szv = szvv = 0;

		for(int i = 0; i < n; i ++) {
			if(m[i]) v[szv++] = i, cn ++;
			else vv[szvv++] = i;
		}

		long long r = 0;
		for(int ii = 0; ii < szvv; ii ++) {
			int i = vv[ii];

			int mn = min(abs(x[i]), abs(y[i]));
			for(int jj = 0; jj < szv; jj ++) {
				int j = v[jj];
				if(m[j] == 1) mn = min(mn, abs(x[i] - x[j]));
				else if(m[j] == 2) mn = min(mn, abs(y[i] - y[j]));
			}

			r += 1ll * mn * w[i];
			if(r > rlt[cn]) return;
		}
		for(int i = cn; i <= n; i ++) {
			if(rlt[i] > r) {
				rlt[i] = r;
			}
			else return;
		}
	}
	else{
		m[c] = 0; DFS(c + 1);
		m[c] = 1; DFS(c + 1);
		m[c] = 2; DFS(c + 1);
	}
}

int id[20], xx[20], yy[20], ww[20];

int main(){
  	srand(time(0));
	scanf("%d", &n);
	for(int i = 0; i < n; i ++){
		scanf("%d%d%lld", &xx[i], &yy[i], &ww[i]);
		id[i] = i;
	}
	random_shuffle(id, id + n);
	for(int i = 0; i < n; i ++) x[i] = xx[id[i]], y[i] = yy[id[i]], w[i] = ww[id[i]];

	memset(m, 0, sizeof(m));
	memset(rlt, 127, sizeof(rlt));

	DFS(0);

	for(int i = 0; i <= n; i ++){
		printf("%lld\n", rlt[i]);
	}
}