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;

long long N;
long long X[20];
long long Y[20];
long long P[20];

#define NO 1
#define YOKO 2
#define TATE 3
long long pat[20];

//   num     , cost
map<long long, long long> memo;

long long xx[20];
long long yy[20];
long long noidx[20];

void cal() {
	long long xxidx = 0;
	long long yyidx = 0;
	long long noidxidx = 0;
	long long sum = 0;

	for(long long i = 0; i < N; i++) {
		if(pat[i] == TATE) {
			xx[xxidx] = X[i];
			xxidx++;
		} else if(pat[i] == YOKO) {
			yy[yyidx] = Y[i];
			yyidx++;
		} else {
			noidx[noidxidx] = i;
			noidxidx++;
		}
	}

	for(long long tmptmp = 0; tmptmp < noidxidx; tmptmp++) {
		long long i = noidx[tmptmp];
		long long tmp = min(abs(X[i]), abs(Y[i]));
		for(long long j = 0; j < xxidx; j++) {
			long long x = xx[j];
			tmp = min(tmp, abs(x - X[i]));
		}
		for(long long j = 0; j < yyidx; j++) {
			long long y = yy[j];
			tmp = min(tmp, abs(y - Y[i]));
		}
		sum += tmp * P[i];
	}

	long long num = xxidx + yyidx;
	memo[num] = min(memo[num], sum);
}

void dfs(long long dep) {
	if(dep == N) {
		cal();
		return;
	}
	pat[dep] = NO;
	dfs(dep+1);
	pat[dep] = YOKO;
	dfs(dep+1);
	pat[dep] = TATE;
	dfs(dep+1);
}


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

	dfs(0);

	for(long long i = 0; i < N + 1; i++) {
		cout << memo[i] << endl;
	}
}