#define DIN freopen("input.txt","r",stdin);
#define DOUT freopen("output.txt","w",stdout);
#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define REP(i,a,b) for(int i=(a);i<=(int)(b);i++)
#define REP_(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
using namespace std;
typedef long long LL;
typedef vector<int> VI;
typedef pair<LL,LL> P;
int read()
{
    int x=0,flag=1; char c=getchar();
    while((c>'9' || c<'0') && c!='-') c=getchar();
    if(c=='-') flag=0,c=getchar();
    while(c<='9' && c>='0') {x=(x<<3)+(x<<1)+c-'0';c=getchar();}
    return flag?x:-x;
}

const int maxn=16;
LL n,x[maxn],y[maxn],p[maxn],ans[maxn],m[maxn];
LL xx[maxn],yy[maxn];
int a[maxn],nx,ny;

void solve(int q)
{
    int tot=0;
    REP(i,1,n)
    {
        a[i]=q%3; q/=3;
        if(a[i]>0) tot++;
    }

    nx=ny=0;
    REP(i,1,n)
        if(a[i]==1) xx[++nx]=x[i];
        else if(a[i]==2) yy[++ny]=y[i];
    REP(i,1,n) m[i]=3e18;
    LL t=0;
    REP(i,1,n)
    {
        REP(j,1,nx) m[i]=min(m[i],abs(xx[j]-x[i]));
        REP(j,1,ny) m[i]=min(m[i],abs(yy[j]-y[i]));
        m[i]=min(m[i],abs(x[i]));
        m[i]=min(m[i],abs(y[i]));
        t+=m[i]*p[i];
        if(t>=ans[tot]) return;
    }
    ans[tot]=min(ans[tot],t);

}

int main()
{
    n=read();
    REP(i,1,n) x[i]=read(),y[i]=read(),p[i]=read();
    REP(i,0,n) ans[i]=3e18;
    int q=1;
    REP(i,1,n) q*=3;
    REP(i,0,q-1) solve(i);
    REP(i,0,n) printf("%lld\n",ans[i]);

    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#define DIN freopen("input.txt","r",stdin);
#define DOUT freopen("output.txt","w",stdout);
#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define REP(i,a,b) for(int i=(a);i<=(int)(b);i++)
#define REP_(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
using namespace std;
typedef long long LL;
typedef vector<int> VI;
typedef pair<LL,LL> P;
int read()
{
    int x=0,flag=1; char c=getchar();
    while((c>'9' || c<'0') && c!='-') c=getchar();
    if(c=='-') flag=0,c=getchar();
    while(c<='9' && c>='0') {x=(x<<3)+(x<<1)+c-'0';c=getchar();}
    return flag?x:-x;
}

const int maxn=16;
LL n,x[maxn],y[maxn],p[maxn],ans[maxn],m[maxn];
LL xx[maxn],yy[maxn];
int a[maxn],nx,ny;

void solve(int q)
{
    int tot=0;
    REP(i,1,n)
    {
        a[i]=q%3; q/=3;
        if(a[i]>0) tot++;
    }

    nx=ny=0;
    REP(i,1,n)
        if(a[i]==1) xx[++nx]=x[i];
        else if(a[i]==2) yy[++ny]=y[i];
    REP(i,1,n) m[i]=3e18;
    LL t=0;
    REP(i,1,n)
    {
        REP(j,1,nx) m[i]=min(m[i],abs(xx[j]-x[i]));
        REP(j,1,ny) m[i]=min(m[i],abs(yy[j]-y[i]));
        m[i]=min(m[i],abs(x[i]));
        m[i]=min(m[i],abs(y[i]));
        t+=m[i]*p[i];
        if(t>=ans[tot]) return;
    }
    ans[tot]=min(ans[tot],t);

}

int main()
{
    n=read();
    REP(i,1,n) x[i]=read(),y[i]=read(),p[i]=read();
    REP(i,0,n) ans[i]=3e18;
    int q=1;
    REP(i,1,n) q*=3;
    REP(i,0,q-1) solve(i);
    REP(i,0,n) printf("%lld\n",ans[i]);

    return 0;
}

```