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 MAXN 200005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n,x[MAXN],y[MAXN],p[MAXN];
ll ans[MAXN];
int distx[15][(1<<15)],disty[15][(1<<15)];
void solve(int mask,int submask)
{
    ll res=0;
    for(int i=0;i<n;i++)
    {
        if(!(mask&(1<<i))) 
        {
            int dist=min(max(x[i],-x[i]),max(y[i],-y[i]));
            dist=min(dist,distx[i][submask]);
            dist=min(dist,disty[i][mask^submask]);
            res+=1LL*dist*p[i];
        }
    }
    ans[__builtin_popcount(mask)]=min(ans[__builtin_popcount(mask)],res);
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d%d%d",&x[i],&y[i],&p[i]);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<(1<<n);j++)
        {
            distx[i][j]=INF;
            for(int k=0;k<n;k++) if(j&(1<<k)) distx[i][j]=min(distx[i][j],max(x[i]-x[k],x[k]-x[i]));
        }
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<(1<<n);j++)
        {
            disty[i][j]=INF;
            for(int k=0;k<n;k++) if(j&(1<<k)) disty[i][j]=min(disty[i][j],max(y[i]-y[k],y[k]-y[i]));
        }
    }
    for(int i=0;i<=n;i++) ans[i]=1000000000000000000LL;
    for(int i=0;i<(1<<n);i++)
    {
        int sub=i;
        do
        {
            solve(i,sub);
            sub=(sub-1)&i;
        }while(sub!=i);
    }
    for(int i=0;i<=n;i++) printf("%lld\n",ans[i]);
    return 0;
}