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;
using ll = long long;

const ll INF = 1e18;

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	int n;
	cin >> n;
	vector<int> x(n), y(n), p(n);
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i] >> p[i];
	}

	vector<vector<int>> dp_x(n, vector<int>(1 << n, 1e9));
	for (int mask = 1; mask < (1 << n); mask++) {
		int one = __builtin_ctz(mask);
		int pmask = mask - (1 << one);

		for (int i = 0; i < n; i++) {
			dp_x[i][mask] = min(dp_x[i][pmask], abs(x[i] - x[one]));
		}
	}

	vector<vector<int>> dp_y(n, vector<int>(1 << n, 1e9));
	for (int mask = 1; mask < (1 << n); mask++) {
		int one = __builtin_ctz(mask);
		int pmask = mask - (1 << one);

		for (int i = 0; i < n; i++) {
			dp_y[i][mask] = min(dp_y[i][pmask], abs(y[i] - y[one]));
		}
	}

	vector<ll> dp(1 << n, INF);
	for (int mask = 0; mask < (1 << n); mask++) {
		vector<int> ones;
		for (int i = 0; i < n; i++) {
			if ((mask >> i) & 1) {
				ones.push_back(i);
			}
		}

		for (int pmask = 0; pmask < (1 << ones.size()); pmask++) {
			int mask_x = 0, mask_y = 0;
			for (int i = 0; i < ones.size(); i++) {
				(((pmask >> i) & 1) ? mask_x : mask_y) += (1 << ones[i]);
			}

			ll res = 0;
			for (int i = 0; i < n; i++) {
				res += (ll)min({abs(x[i]), abs(y[i]), dp_x[i][mask_x], dp_y[i][mask_y]}) * p[i];
			}

			dp[mask] = min(dp[mask], res);
		}
	}

	vector<ll> ans(n + 1, INF);
	for (int mask = 0; mask < (1 << n); mask++) {
		int cnt = __builtin_popcount(mask);
		ans[cnt] = min(ans[cnt], dp[mask]);
	}

	for (auto el : ans) {
		cout << el << '\n';
	}
}