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 <vector>
#include <algorithm>
using namespace std;
using ll = long long;
const ll INF = 1e18;

const int N = 15;
ll dist[N][2][1 << N];

ll xs[N];
ll ys[N];
ll ps[N];

ll res[N+1];

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

	// Optimal to put straight through at least one city
	//	PROOF: assume not, assume two use same railway, move it towards the more populous using side

	int n;
	cin >> n;

	for (int i = 0; i < n; ++i) cin >> xs[i] >> ys[i] >> ps[i];

	for (int a = 0; a < n; ++a) {
		ll base = ps[a] * min(abs(xs[a]), abs(ys[a]));
		for (int m = 0; m < (1 << n); ++m) {
			dist[a][0][m] = base;
			dist[a][1][m] = base;
			for (int b = 0; b < n; ++b) {
				if (m & (1 << b)) {
					dist[a][0][m] = min(dist[a][0][m], ps[a] * abs(xs[a] - xs[b]));
					dist[a][1][m] = min(dist[a][1][m], ps[a] * abs(ys[a] - ys[b]));
				}
			}
		}
	}

	for (int i = 0; i <= n; ++i) res[i] = INF;
	
	for (int m = 0; m < (1 << n); ++m) {
		ll off = INF;
		for (int k = m;; k = (k-1) & m) {
			ll cur = 0;
			for (int a = 0; a < n; ++a) {
				cur += min(dist[a][0][m & k], dist[a][1][m & (~k)]);
			}
			off = min(off, cur);

			if (k == 0) break;
		}
		int i = __builtin_popcount(m);
		res[i] = min(res[i], off);
	}
	for (int i = 0; i <= n; ++i) cout << res[i] << ' '; cout << '\n';
}