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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

constexpr int64_t INF = 1e18;
constexpr int MAX = 15;
int n;
int64_t x[MAX], y[MAX], p[MAX];
int64_t dpx[MAX][1 << MAX];
int64_t dpy[MAX][1 << MAX];
int64_t ans[MAX + 1];

void init() {
	for(int i = 0; i < MAX; ++i) {
		ans[i] = INF;
		for(int j = 0; j < (1 << MAX); ++j) {
			dpx[i][j] = abs(x[i]);
			dpy[i][j] = abs(y[i]);
		}
	}
	for(int j = 0; j < (1 << MAX); ++j) {
		for(int k = 0; k < MAX; ++k) {
			if(j >> k & 1) {
				for(int i = 0; i < MAX; ++i) {
					dpx[i][j] = min(dpx[i][j], abs(x[i] - x[k]));
					dpy[i][j] = min(dpy[i][j], abs(y[i] - y[k]));
				}
			}
		}
	}
}

int main() {
	cin >> n;
	for(int i = 0; i < n; ++i) cin >> x[i] >> y[i] >> p[i];
	init();
	for(int mask = 0; mask < pow(3, n); ++mask) {
		int msk = mask;
		int mskx = 0, msky = 0;
		int cnt = 0;
		for(int i = 0; i < n; ++i) {
			if(msk % 3 == 1) {
				mskx |= (1 << i);
				++cnt;
			}
			if(msk % 3 == 2) {
				msky |= (1 << i);
				++cnt;
			}
			msk /= 3;
		}
		int64_t sum = 0;
		for(int i = 0; i < n; ++i) {
			sum += min(dpx[i][mskx], dpy[i][msky]) * p[i];
		}
		ans[cnt] = min(ans[cnt], sum);
	}
	for(int i = 0; i <= n; ++i) {
		cout << ans[i] << '\n';
	}
	return 0;
}