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 <stdio.h>
int n;
long long t, s, u, v, q, f;
int a[15];
long long x[15];
long long y[15];
long long p[15];
long long r[20];
long long rr[20];
long long b[1024 * 32][15];
long long c[1024 * 32][15];
long long ans[16];
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lld%lld%lld", x + i, y + i, p + i);
	}
	for (int i = 0; i <= n; i++) {
		ans[i] = 150000;
		ans[i] *= 1000000;
	}
	r[0] = 1;
	rr[0] = 1;
	for (int i = 0; i <= 15; i++) {
		r[i + 1] = r[i] * 3;
		rr[i + 1] = rr[i] * 2;
	}
	for (int i = 0; i < rr[n]; i++) {
		t = i;
		for (int j = 0; j < n; j++) {
			a[j] = t % 2;
			t /= 2;
		}
		for (int j = 0; j < n; j++) {
			s = x[j];
			if (x[j] < 0)s = -x[j];
			for (int k = 0; k < n; k++) {
				if (a[k]) {
					u = x[j] - x[k];
					if (u < 0)u = -u;
					if (s > u)s = u;
				}
			}
			b[i][j] = s * p[j];
		}
		for (int j = 0; j < n; j++) {
			s = y[j];
			if (y[j] < 0)s = -y[j];
			for (int k = 0; k < n; k++) {
				if (a[k]) {
					u = y[j] - y[k];
					if (u < 0)u = -u;
					if (s > u)s = u;
				}
			}
			c[i][j] = s * p[j];
		}
	}
	for (int i = 0; i < r[n]; i++) {
		t = i;
		s = 0;
		u = 0;
		q = 0;
		v = 0;
		for (int j = n - 1; j >= 0; j--) {
			a[j] = t % 3;
			if (a[j] == 1)s++;
			if (a[j] == 2)u++;
			t /= 3;
			s *= 2;
			u *= 2;
			if (a[j])q++;
		}
		s /= 2;
		u /= 2;
		for (int j = 0; j < n; j++) {
			if (!a[j]) {
				f = b[s][j];
				if (f > c[u][j])f = c[u][j];
				v += f;
			}
		}
		if (ans[q] > v)ans[q] = v;
	}
	for (int i = 0; i <= n; i++) {
		printf("%lld\n", ans[i]);
	}
}