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;
template<typename T>
inline void read(T &x)
{
	x=0;
	int f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9')
	{
		if(ch=='-') f=-1;
		ch=getchar();
}
	while(ch>='0'&&ch<='9') x=x*10+(ch^48),ch=getchar();
	x*=f;
	return;
}	
template<typename T>
void write(T x)
{
	if(x<0) putchar('-'),x=-x;
	if(x>=10) write(x/10);
	putchar(x%10+'0');
	return;
}
const int MAXN=20;
const long long inf=1e18+7;
int n;
struct node{
    int x,y;
    long long p;
}a[MAXN];
long long ans[MAXN];
int dis[MAXN];
void dfs(int step,int cnt,long long tot)
{
    if(step==n+1)
    {
        ans[cnt]=min(ans[cnt],tot);
        return;
    }
    dfs(step+1,cnt,tot);
    int tmp[MAXN];
    long long ttmp=0;
    for(int i=1;i<=n;++i) tmp[i]=dis[i];
    for(int i=1;i<=n;++i) dis[i]=min(dis[i],abs(a[i].y-a[step].y)),ttmp+=a[i].p*dis[i];
    dfs(step+1,cnt+1,ttmp);
    ttmp=0;
    for(int i=1;i<=n;++i) dis[i]=tmp[i];
    for(int i=1;i<=n;++i) dis[i]=min(dis[i],abs(a[i].x-a[step].x)),ttmp+=a[i].p*dis[i];
    dfs(step+1,cnt+1,ttmp);
    for(int i=1;i<=n;++i) dis[i]=tmp[i];
}
int main()
{
    read(n);
    for(int i=1;i<=n;++i) read(a[i].x),read(a[i].y),read(a[i].p);
    long long tmp=0;
    for(int i=0;i<=n;++i) ans[i]=inf,dis[i]=min(abs(a[i].x),abs(a[i].y)),tmp+=a[i].p*dis[i];
    dfs(1,0,tmp);
    for(int i=0;i<=n;++i) write(ans[i]),putchar('\n');
	return 0;
}