#include<bits/stdc++.h>
#define mod 1000000000
#define infll 1000000000000000000LL
#define pb push_back
#define sz(v) (v).size()
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
#define mp make_pair
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define REP(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
using namespace std;
template <typename T> void chmax(T &x,T y){x=x>y?x:y;return;}
template <typename T> void chmin(T &x,T y){x=x<y?x:y;return;}
typedef long long ll;
typedef unsigned long long ull;

int n,x[25],y[25],p[25],res[25][25];
ll ans[25];

void dfs(int i,int cnt)
{
	if(i>n)
	{
		ll ret=0;
		rep(j,1,n) ret+=ll(res[i][j])*ll(p[j]);
		chmin(ans[cnt],ret);
		return;
	}
	rep(j,1,n) res[i+1][j]=res[i][j];
	dfs(i+1,cnt);
	rep(j,1,n) res[i+1][j]=min(res[i][j],abs(x[i]-x[j]));
	dfs(i+1,cnt+1);
	rep(j,1,n) res[i+1][j]=min(res[i][j],abs(y[i]-y[j]));
	dfs(i+1,cnt+1);
	return;
}

int main()
{
	scanf("%d",&n);
	rep(i,1,n) scanf("%d%d%d",&x[i],&y[i],&p[i]);
	rep(i,1,n) res[1][i]=min(abs(x[i]),abs(y[i]));
	rep(i,0,n) ans[i]=infll;
	dfs(1,0);
	rep(i,0,n) printf("%lld\n",ans[i]);
	return 0;
}
### Prompt

Create a solution in cpp for the following 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

### Response

```cpp
#include<bits/stdc++.h>
#define mod 1000000000
#define infll 1000000000000000000LL
#define pb push_back
#define sz(v) (v).size()
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
#define mp make_pair
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define REP(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
using namespace std;
template <typename T> void chmax(T &x,T y){x=x>y?x:y;return;}
template <typename T> void chmin(T &x,T y){x=x<y?x:y;return;}
typedef long long ll;
typedef unsigned long long ull;

int n,x[25],y[25],p[25],res[25][25];
ll ans[25];

void dfs(int i,int cnt)
{
	if(i>n)
	{
		ll ret=0;
		rep(j,1,n) ret+=ll(res[i][j])*ll(p[j]);
		chmin(ans[cnt],ret);
		return;
	}
	rep(j,1,n) res[i+1][j]=res[i][j];
	dfs(i+1,cnt);
	rep(j,1,n) res[i+1][j]=min(res[i][j],abs(x[i]-x[j]));
	dfs(i+1,cnt+1);
	rep(j,1,n) res[i+1][j]=min(res[i][j],abs(y[i]-y[j]));
	dfs(i+1,cnt+1);
	return;
}

int main()
{
	scanf("%d",&n);
	rep(i,1,n) scanf("%d%d%d",&x[i],&y[i],&p[i]);
	rep(i,1,n) res[1][i]=min(abs(x[i]),abs(y[i]));
	rep(i,0,n) ans[i]=infll;
	dfs(1,0);
	rep(i,0,n) printf("%lld\n",ans[i]);
	return 0;
}
```