#include <stdio.h>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>

using ll = long long int;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

int st[20],d[20],n,x[20],y[20];
ll ans[20],p[20],calc;

void solve(int i, int k){
	int pr[20],j,t;
	if(i == n){
		calc = 0ll;
		for(j=0; j<n; ++j){
			calc += p[j]*d[j];
		}
		ans[k] = std::min(ans[k],calc);
		return;
	}
	for(j=0; j<n; ++j){
		pr[j] = d[j];
	}
	for(t=0; t<3; ++t){
		if(t==1){
			for(j=0; j<n; ++j){
				d[j] = std::min(d[j],abs(x[i]-x[j]));
			}
			solve(i+1,k+1);
		}else if(t==2){
			for(j=0; j<n; ++j){
				d[j] = std::min(d[j],abs(y[i]-y[j]));
			}
			solve(i+1,k+1);
		}else{
			solve(i+1,k);
		}
		for(j=0; j<n; ++j){
			d[j] = pr[j];
		}
	}
}

int main(void){
	int i;
	ll calc=0ll;
	std::fill(ans,ans+20,INFLL);
	std::cin >> n;
	for(i=0; i<n; ++i){
		std::cin >> x[i] >> y[i] >> p[i];
		d[i] = std::min(abs(x[i]),abs(y[i]));
	}
	solve(0,0);
	for(i=0; i<=n; ++i){
		std::cout << ans[i] << std::endl;
	}
	return 0;
}

### Prompt

In cpp, your task is to solve 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 <stdio.h>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>

using ll = long long int;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

int st[20],d[20],n,x[20],y[20];
ll ans[20],p[20],calc;

void solve(int i, int k){
	int pr[20],j,t;
	if(i == n){
		calc = 0ll;
		for(j=0; j<n; ++j){
			calc += p[j]*d[j];
		}
		ans[k] = std::min(ans[k],calc);
		return;
	}
	for(j=0; j<n; ++j){
		pr[j] = d[j];
	}
	for(t=0; t<3; ++t){
		if(t==1){
			for(j=0; j<n; ++j){
				d[j] = std::min(d[j],abs(x[i]-x[j]));
			}
			solve(i+1,k+1);
		}else if(t==2){
			for(j=0; j<n; ++j){
				d[j] = std::min(d[j],abs(y[i]-y[j]));
			}
			solve(i+1,k+1);
		}else{
			solve(i+1,k);
		}
		for(j=0; j<n; ++j){
			d[j] = pr[j];
		}
	}
}

int main(void){
	int i;
	ll calc=0ll;
	std::fill(ans,ans+20,INFLL);
	std::cin >> n;
	for(i=0; i<n; ++i){
		std::cin >> x[i] >> y[i] >> p[i];
		d[i] = std::min(abs(x[i]),abs(y[i]));
	}
	solve(0,0);
	for(i=0; i<=n; ++i){
		std::cout << ans[i] << std::endl;
	}
	return 0;
}

```