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 <vector>
#include <iostream>
#include <algorithm>
std::vector<long long> solve(int N, std::vector<int> X, std::vector<int> Y, std::vector<int> P) {
	const long long inf = 1LL << 62;
	int mul = 1;
	for (int i = 0; i < N; ++i) {
		mul *= 3;
	}
	std::vector<std::vector<int> > xsel(1 << N, std::vector<int>(N));
	std::vector<std::vector<int> > ysel(1 << N, std::vector<int>(N));
	for (int i = 0; i < 1 << N; ++i) {
		for (int j = 0; j < N; ++j) {
			xsel[i][j] = std::abs(X[j]);
			ysel[i][j] = std::abs(Y[j]);
			for (int k = 0; k < N; ++k) {
				if ((i >> k) & 1) {
					xsel[i][j] = std::min(xsel[i][j], std::abs(X[j] - X[k]));
					ysel[i][j] = std::min(ysel[i][j], std::abs(Y[j] - Y[k]));
				}
			}
		}
	}
	std::vector<long long> ans(N + 1, inf);
	for (int i = 0; i < 1 << N; ++i) {
		int cnt = 0;
		for (int j = 0; j < N; ++j) {
			if ((i >> j) & 1) ++cnt;
		}
		for (int j = i; j >= 0; --j) {
			j &= i;
			long long cost = 0;
			for (int k = 0; k < N; ++k) {
				if (!((i >> k) & 1)) {
					cost += 1LL * std::min({ xsel[j][k], ysel[i - j][k] }) * P[k];
				}
			}
			ans[cnt] = std::min(ans[cnt], cost);
		}
	}
	return ans;
}
int main() {
	int N;
	std::cin >> N;
	std::vector<int> X(N), Y(N), P(N);
	for (int i = 0; i < N; ++i) {
		std::cin >> X[i] >> Y[i] >> P[i];
	}
	std::vector<long long> ans = solve(N, X, Y, P);
	for (int i = 0; i <= N; ++i) {
		std::cout << ans[i] << std::endl;
	}
	return 0;
}