#include<iostream>
#include<cmath>
#define x first
#define y second
using namespace std;
const int N = 16, M = (1 << 15);
const long long INF = 0x3f3f3f3f3f3f3f3f;
int n, log_2[1 << N];
long long dx[M][N], dy[M][N], x[N], y[N], p[N], ans;

void dfs(int a, int b, int maska, int maskb, int posi)
{
	if(a == 0 && b == 0)
	{
		long long tmp = 0;
		for(int i = 0; i < n; ++i)
			tmp += min(dx[maska][i], dy[maskb][i]);
		ans = min(ans, tmp);
		return;
	}
	if(n - posi < a + b) return;
	if(a > 0) dfs(a-1, b, maska | 1 << posi, maskb, posi + 1);
	if(b > 0) dfs(a, b-1, maska, maskb | 1 << posi, posi + 1);
	dfs(a, b, maska, maskb, posi + 1);
}
int main()
{
	cin >> n;
	for(int i = 0; i < n; ++i) cin >> x[i] >> y[i] >> p[i];
	for(int i = 0; i < n; ++i)
		dx[0][i] = abs(y[i] * p[i]), dy[0][i] = abs(x[i] * p[i]), log_2[1<<i] = i;
	for(int i = 1; i < 1 << n; ++i)
	{
		for(int j = 0; j < n; ++j)
		{
			long long tmpx = abs(y[j] - y[log_2[i & -i]]),
					  tmpy = abs(x[j] - x[log_2[i & -i]]);
			dx[i][j] = min(dx[i & (i - 1)][j], tmpx * p[j]);
			dy[i][j] = min(dy[i & (i - 1)][j], tmpy * p[j]);
		}
	}
//	for(int i = 0; i < 1 << n; ++i)
//		for(int j = 0; j < n; ++j)
//			printf("dx[%d][%d] = %lld, dj[%d][%d] = %lld\n", i, j, dx[i][j], i, j, dy[i][j]);
//
	for(int k = 0; k < n; ++k)
	{
		ans = INF;
		for(int a = 0; a <= k; ++a)
			dfs(a, k - a, 0, 0, 0);
		cout << ans << endl;
	}
	cout << 0 << endl;
	return 0;
}

### Prompt

Generate 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
#include<iostream>
#include<cmath>
#define x first
#define y second
using namespace std;
const int N = 16, M = (1 << 15);
const long long INF = 0x3f3f3f3f3f3f3f3f;
int n, log_2[1 << N];
long long dx[M][N], dy[M][N], x[N], y[N], p[N], ans;

void dfs(int a, int b, int maska, int maskb, int posi)
{
	if(a == 0 && b == 0)
	{
		long long tmp = 0;
		for(int i = 0; i < n; ++i)
			tmp += min(dx[maska][i], dy[maskb][i]);
		ans = min(ans, tmp);
		return;
	}
	if(n - posi < a + b) return;
	if(a > 0) dfs(a-1, b, maska | 1 << posi, maskb, posi + 1);
	if(b > 0) dfs(a, b-1, maska, maskb | 1 << posi, posi + 1);
	dfs(a, b, maska, maskb, posi + 1);
}
int main()
{
	cin >> n;
	for(int i = 0; i < n; ++i) cin >> x[i] >> y[i] >> p[i];
	for(int i = 0; i < n; ++i)
		dx[0][i] = abs(y[i] * p[i]), dy[0][i] = abs(x[i] * p[i]), log_2[1<<i] = i;
	for(int i = 1; i < 1 << n; ++i)
	{
		for(int j = 0; j < n; ++j)
		{
			long long tmpx = abs(y[j] - y[log_2[i & -i]]),
					  tmpy = abs(x[j] - x[log_2[i & -i]]);
			dx[i][j] = min(dx[i & (i - 1)][j], tmpx * p[j]);
			dy[i][j] = min(dy[i & (i - 1)][j], tmpy * p[j]);
		}
	}
//	for(int i = 0; i < 1 << n; ++i)
//		for(int j = 0; j < n; ++j)
//			printf("dx[%d][%d] = %lld, dj[%d][%d] = %lld\n", i, j, dx[i][j], i, j, dy[i][j]);
//
	for(int k = 0; k < n; ++k)
	{
		ans = INF;
		for(int a = 0; a <= k; ++a)
			dfs(a, k - a, 0, 0, 0);
		cout << ans << endl;
	}
	cout << 0 << endl;
	return 0;
}

```