#pragma GCC optimize(2)

#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstring>

#include <set>
#include <queue>
#include <complex>
#include <algorithm>

int read(void){
	int res = 0, sign = 1;
	char ch = std::getchar();
	while(!std::isdigit(ch)){
		if(ch == '-')
			sign = -1;
		ch = std::getchar();
	}
	while(std::isdigit(ch))
		res = res * 10 + ch - 48, ch = std::getchar();
	return sign * res;
}

template<class Tp>
	void write(const Tp &x){
		if(x > 9)
			write(x / 10);
		std::putchar(x % 10 + 48);
	}

typedef long long ll;
typedef unsigned long long ull;
typedef double rl;
typedef long double RL;

const rl PI = 3.1415926535898, E = 2.71828182845904;

const int MAXN = 18;

ll ans[MAXN];
int dist[MAXN];

int n;
int x[MAXN], y[MAXN], p[MAXN];
int in[MAXN];

int main(){
	n = read();
	for(int i = 1; i <= n; ++i)
		x[i] = read(), y[i] = read(), p[i] = read();
	for(int i = 0; i <= n; ++i)
		ans[i] = 1e18;
	for(int i = 0; i < (1 << n); ++i){
		int cnt = 0;
		for(int j = 1; j <= n; ++j)
			if(i & (1 << (j - 1)))
				in[++cnt] = j;
		for(int j = 0; j < (1 << cnt); ++j){
			for(int p = 1; p <= n; ++p)
				dist[p] = std::min(int(std::abs(x[p])), int(std::abs(y[p])));
			for(int k = 1; k <= cnt; ++k)
				if(j & (1 << (k - 1)))
					for(int p = 1; p <= n; ++p)
						dist[p] = std::min(dist[p], int(std::abs(x[p] - x[in[k]])));
				else
					for(int p = 1; p <= n; ++p)
						dist[p] = std::min(dist[p], int(std::abs(y[p] - y[in[k]])));
			ll s = 0ll;
			for(int j = 1; j <= n; ++j)
				s += (ll)dist[j] * p[j];
			ans[cnt] = std::min(ans[cnt], s);
		}
	}
	for(int i = 0; i <= n; ++i)
		std::printf("%lld\n", ans[i]);
	return 0;
}
### Prompt

Your challenge is to write a Cpp solution to 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
#pragma GCC optimize(2)

#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstring>

#include <set>
#include <queue>
#include <complex>
#include <algorithm>

int read(void){
	int res = 0, sign = 1;
	char ch = std::getchar();
	while(!std::isdigit(ch)){
		if(ch == '-')
			sign = -1;
		ch = std::getchar();
	}
	while(std::isdigit(ch))
		res = res * 10 + ch - 48, ch = std::getchar();
	return sign * res;
}

template<class Tp>
	void write(const Tp &x){
		if(x > 9)
			write(x / 10);
		std::putchar(x % 10 + 48);
	}

typedef long long ll;
typedef unsigned long long ull;
typedef double rl;
typedef long double RL;

const rl PI = 3.1415926535898, E = 2.71828182845904;

const int MAXN = 18;

ll ans[MAXN];
int dist[MAXN];

int n;
int x[MAXN], y[MAXN], p[MAXN];
int in[MAXN];

int main(){
	n = read();
	for(int i = 1; i <= n; ++i)
		x[i] = read(), y[i] = read(), p[i] = read();
	for(int i = 0; i <= n; ++i)
		ans[i] = 1e18;
	for(int i = 0; i < (1 << n); ++i){
		int cnt = 0;
		for(int j = 1; j <= n; ++j)
			if(i & (1 << (j - 1)))
				in[++cnt] = j;
		for(int j = 0; j < (1 << cnt); ++j){
			for(int p = 1; p <= n; ++p)
				dist[p] = std::min(int(std::abs(x[p])), int(std::abs(y[p])));
			for(int k = 1; k <= cnt; ++k)
				if(j & (1 << (k - 1)))
					for(int p = 1; p <= n; ++p)
						dist[p] = std::min(dist[p], int(std::abs(x[p] - x[in[k]])));
				else
					for(int p = 1; p <= n; ++p)
						dist[p] = std::min(dist[p], int(std::abs(y[p] - y[in[k]])));
			ll s = 0ll;
			for(int j = 1; j <= n; ++j)
				s += (ll)dist[j] * p[j];
			ans[cnt] = std::min(ans[cnt], s);
		}
	}
	for(int i = 0; i <= n; ++i)
		std::printf("%lld\n", ans[i]);
	return 0;
}
```