#include<stdio.h>
#pragma GCC optimize(2)
short x[16],y[16];
int p[16];
long long ans[16];
short mi[16],m1[16];
short px[16][16],py[16][16];
int ppk,st;
short cnt,now;
int main(void)
{
 
	char n;
	scanf("%hd",&n);
	for(char i=0;i<n;i++) scanf("%hd%hd%d",&x[i],&y[i],&p[i]);
	int ma=1;
	for(char i=0;i<=n;++i) ans[i]=1000000000000000;
	for(char i=0;i<n;++i) ma*=3;
	for(char i=0;i<n;++i)
	{
		short x1=x[i]>0?x[i]:-x[i],y1=y[i]>0?y[i]:-y[i];
		m1[i]=x1>y1?y1:x1;
	}
	for(char i=0;i<n;++i)
	{
		for(char j=0;j<n;++j)
		{
			px[i][j]=x[i]-x[j];
			py[i][j]=y[i]-y[j];
			if(px[i][j]<0) px[i][j]=-px[i][j];
			if(py[i][j]<0) py[i][j]=-py[i][j];
		}
	}
	for(int zhu=0;zhu<ma;++zhu)
	{
		ppk=zhu,cnt=n,st=ma/3;
		for(char i=0;i<n;i++) mi[i]=m1[i];
		for(char i=0;i<n;i++,st/=3)
		{
			now=ppk/st;
			switch(now)
			{
				case 2:for(char j=0;j<n;++j) if(mi[j]>px[i][j]) mi[j]=px[i][j];break;
				case 1:for(char j=0;j<n;++j) if(mi[j]>py[i][j]) mi[j]=py[i][j];break;
				case 0:cnt--;
			}
			ppk-=st*now;
		}
		long long sum=0;
		for(char i=0;i<n;++i) sum+=(long long)mi[i]*p[i];
		if(ans[cnt]>sum) ans[cnt]=sum;
	}
	for(char i=0;i<=n;i++) printf("%lld\n",ans[i]);
}

### 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
#include<stdio.h>
#pragma GCC optimize(2)
short x[16],y[16];
int p[16];
long long ans[16];
short mi[16],m1[16];
short px[16][16],py[16][16];
int ppk,st;
short cnt,now;
int main(void)
{
 
	char n;
	scanf("%hd",&n);
	for(char i=0;i<n;i++) scanf("%hd%hd%d",&x[i],&y[i],&p[i]);
	int ma=1;
	for(char i=0;i<=n;++i) ans[i]=1000000000000000;
	for(char i=0;i<n;++i) ma*=3;
	for(char i=0;i<n;++i)
	{
		short x1=x[i]>0?x[i]:-x[i],y1=y[i]>0?y[i]:-y[i];
		m1[i]=x1>y1?y1:x1;
	}
	for(char i=0;i<n;++i)
	{
		for(char j=0;j<n;++j)
		{
			px[i][j]=x[i]-x[j];
			py[i][j]=y[i]-y[j];
			if(px[i][j]<0) px[i][j]=-px[i][j];
			if(py[i][j]<0) py[i][j]=-py[i][j];
		}
	}
	for(int zhu=0;zhu<ma;++zhu)
	{
		ppk=zhu,cnt=n,st=ma/3;
		for(char i=0;i<n;i++) mi[i]=m1[i];
		for(char i=0;i<n;i++,st/=3)
		{
			now=ppk/st;
			switch(now)
			{
				case 2:for(char j=0;j<n;++j) if(mi[j]>px[i][j]) mi[j]=px[i][j];break;
				case 1:for(char j=0;j<n;++j) if(mi[j]>py[i][j]) mi[j]=py[i][j];break;
				case 0:cnt--;
			}
			ppk-=st*now;
		}
		long long sum=0;
		for(char i=0;i<n;++i) sum+=(long long)mi[i]*p[i];
		if(ans[cnt]>sum) ans[cnt]=sum;
	}
	for(char i=0;i<=n;i++) printf("%lld\n",ans[i]);
}

```