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>
#include <limits>

using namespace std;

int main()
{
	int N;
	cin >> N;
	vector<long long> X(N), Y(N);
	vector<long long> P(N);

	for (int i = 0; i < N; ++i) {
		cin >> X[i] >> Y[i] >> P[i];
	}

	vector<vector<long long>> h(N, vector<long long>(1 << N));
	vector<vector<long long>> v(N, vector<long long>(1 << N));

	for (int i = 0; i < N; ++i) {
		for (int mask = 0; mask < (1 << N); ++mask) {
			h[i][mask] = P[i] * abs(Y[i]);
			v[i][mask] = P[i] * abs(X[i]);
			for (int k = 0; k < N; ++k) {
				if (mask & (1 << k)) {
					h[i][mask] = min(h[i][mask], P[i] * abs(Y[i] - Y[k]));
					v[i][mask] = min(v[i][mask], P[i] * abs(X[i] - X[k]));
				}
			}
		}
	}

	vector<int> p3(N + 1);
	p3[0] = 1;
	for (int i = 1; i <= N; ++i) {
		p3[i] = 3 * p3[i - 1];
	}

	vector<long long> ans(N + 1, numeric_limits<long long>::max());

	for (int mask = 0; mask < p3[N]; ++mask) {
		int mask_h = 0;
		int mask_v = 0;
		int k = N;
		for (int i = 0; i < N; ++i) {
			int t = (mask / p3[i]) % 3;
			if (t == 1) {
				mask_h |= 1 << i;
			}
			else if (t == 2) {
				mask_v |= 1 << i;
			}
			else {
				--k;
			}
		}

		long long result = 0;
		for (int i = 0; i < N; ++i) {
			result += min(h[i][mask_h], v[i][mask_v]);
		}
		ans[k] = min(ans[k], result);
	}

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