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;
#define ll long long
#define all(V) V.begin(),V.end()
const ll MOD = 1000000007;
 
ll X[18], Y[18], P[18], N;
ll R[18];
 
void dfs(vector<int>& XX, vector<int>& YY, int c, int k) {
	if (c == N) {
		ll xd = MOD, yd = MOD, S = 0;
		for (int i = 0;i < N;i++) {
			xd = MOD, yd = MOD;
			for (ll x : XX) {
				xd = min(xd, abs(x - X[i]));
			}
			for (ll y : YY) {
				yd = min(yd, abs(y - Y[i]));
			}
			S += min(xd, yd) * P[i];
		}
		R[k] = min(R[k], S);
		return;
	}
 
	XX.push_back(X[c]);
	dfs(XX, YY, c + 1, k + 1);
	XX.pop_back();
 
	YY.push_back(Y[c]);
	dfs(XX, YY, c + 1, k + 1);
	YY.pop_back();
 
	dfs(XX, YY, c + 1, k);
}
 
int main() {
	cin >> N;
 
	for (int i = 0;i < N;i++) {
		cin >> X[i] >> Y[i] >> P[i];
		R[i] = INT64_MAX;
	}
	R[N] = INT64_MAX;
	vector<int> BX = { 0 };
	vector<int> BY = { 0 };
	dfs(BX, BY, 0, 0);
	
	for (int i = 0;i <= N;i++) {
		cout << R[i] << endl;
	}
}