#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<ll> f(int N, const vector<vector<ll>>& xx, int pp) {
	vector<int> idx;
	for (int i = 0; i < N; ++ i) if ((pp>>i)&1) idx.push_back(i);
	int n = idx.size();
	vector<ll> res(n+1, 1LL<<62);
	for (int p = 0; p < (1<<n); ++ p) {
		int cnt = __builtin_popcount(p);
		int q = 0;
		for (int i = 0; i < n; ++ i) if ((p>>i)&1) q |= 1<<idx[i];
		ll d = 0;
		for (int i : idx) {
			d += xx[i][q];
		}
		res[cnt] = min(res[cnt], d);
	}
	return res;
}
ll g(const vector<ll>& x, int pp, int i) {
	ll dd = abs(x[i]);
	for (unsigned j = 0; j < x.size(); ++ j) if ((pp>>j)&1) {
		dd = min(dd, abs(x[i]-x[j]));
	}
	return dd;
}
int main() {
	int N;
	cin >> N;
	vector<ll> X(N), Y(N), P(N);
	for (int i = 0; i < N; ++ i) cin >> X[i] >> Y[i] >> P[i];
	vector<vector<ll>> xx(N, vector<ll>(1<<N)), yy(N, vector<ll>(1<<N));
	for (int i = 0; i < N; ++ i) {
		for (int p = 0; p < (1<<N); ++ p) {
			xx[i][p] = g(X, p, i) * P[i];
			yy[i][p] = g(Y, p, i) * P[i];
		}
	}
	vector<ll> ans(N+1, 1LL<<62);
	for (int p = 0; p < (1<<N); ++ p) {
		auto ax = f(N, xx, p);
		auto ay = f(N, yy, ((1<<N)-1)&~p);
		for (int i = 0; i < (int)ax.size(); ++ i) {
			for (int j = 0; j < (int)ay.size() && i+j <= N; ++ j) {
				ans[i+j] = min(ans[i+j], ax[i] + ay[j]);
			}
		}
	}
	for (auto x : ans) cout << x << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
using namespace std;
typedef long long ll;
vector<ll> f(int N, const vector<vector<ll>>& xx, int pp) {
	vector<int> idx;
	for (int i = 0; i < N; ++ i) if ((pp>>i)&1) idx.push_back(i);
	int n = idx.size();
	vector<ll> res(n+1, 1LL<<62);
	for (int p = 0; p < (1<<n); ++ p) {
		int cnt = __builtin_popcount(p);
		int q = 0;
		for (int i = 0; i < n; ++ i) if ((p>>i)&1) q |= 1<<idx[i];
		ll d = 0;
		for (int i : idx) {
			d += xx[i][q];
		}
		res[cnt] = min(res[cnt], d);
	}
	return res;
}
ll g(const vector<ll>& x, int pp, int i) {
	ll dd = abs(x[i]);
	for (unsigned j = 0; j < x.size(); ++ j) if ((pp>>j)&1) {
		dd = min(dd, abs(x[i]-x[j]));
	}
	return dd;
}
int main() {
	int N;
	cin >> N;
	vector<ll> X(N), Y(N), P(N);
	for (int i = 0; i < N; ++ i) cin >> X[i] >> Y[i] >> P[i];
	vector<vector<ll>> xx(N, vector<ll>(1<<N)), yy(N, vector<ll>(1<<N));
	for (int i = 0; i < N; ++ i) {
		for (int p = 0; p < (1<<N); ++ p) {
			xx[i][p] = g(X, p, i) * P[i];
			yy[i][p] = g(Y, p, i) * P[i];
		}
	}
	vector<ll> ans(N+1, 1LL<<62);
	for (int p = 0; p < (1<<N); ++ p) {
		auto ax = f(N, xx, p);
		auto ay = f(N, yy, ((1<<N)-1)&~p);
		for (int i = 0; i < (int)ax.size(); ++ i) {
			for (int j = 0; j < (int)ay.size() && i+j <= N; ++ j) {
				ans[i+j] = min(ans[i+j], ax[i] + ay[j]);
			}
		}
	}
	for (auto x : ans) cout << x << endl;
}

```