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>

#define rep2(x,fr,to) for(int x=(fr);x<(to);x++)
#define rep(x,to) for(int x=0;x<(to);x++)
#define repr(x,fr,to) for(int x=(fr);x>=(to);x--)
#define all(c) c.begin(),c.end()
#define sz(v) (int)v.size()

using namespace std;
typedef long long  ll; typedef vector<int> VI; typedef pair<ll,ll> pii; typedef vector<ll> VL; const int MD = 1e9 + 7;
void dbg(){cerr<<"\n";} template <class F,class ...S> void dbg(const F& f, const S&...s){cerr <<f <<": "; dbg(s...);}

int main()
{
	cin.tie(0); ios_base::sync_with_stdio(false);
	int n;
	cin >>n;
	array<ll, 18> x, y, p;
	rep(i, n) cin >>x[i] >>y[i] >>p[i];
	const int pj = 3;
	array<ll, 18> ans;
	ans.fill(1LL<<55);
	int mmx = 1; rep(i, n) mmx*=3;
	rep(i, mmx){
		array<ll, 18> hl, xx, yy;
		int ch=0, cx=0, cy=0;
		for(int j=0, wki=i; j<n; j++, wki/=pj){
			int nw = wki%pj;
			if(nw==0) hl[ch++] = j;
			else if(nw==1) xx[cx++]=x[j];
			else if(nw==2) yy[cy++]=y[j];
		}
		ll sm = 0;
		rep(j, ch){
			auto z = hl[j];
			ll cnd = min(abs(x[z]), abs(y[z]));
			
			rep(k, cx) cnd = min<ll>(cnd, abs(xx[k] - x[z]));
			rep(k, cy) cnd = min<ll>(cnd, abs(yy[k] - y[z]));
			sm += cnd * p[z];
		}
		ans[cx+cy] = min(ans[cx+cy], sm);
	}
	
	rep(i, n+1) cout <<ans[i] <<"\n";
}