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 <cstdio>
#include <cmath>
#include <limits>
#include <algorithm>
using namespace std;
using i64 = long long;
int n, i;
int x[20], y[20], p[20], z[20];
i64 ans[20];
char b[20], s[20], c[20];

void ct(int j)
{
    i64 result = 0;
    for (int i = 1; i <= n; ++i)
    {
        if (b[i] != 0)
            continue;
        int t = z[i];
        for (int u = 0; u < j; ++u)
            if (c[u] == 1)
                t = min(t, abs(x[i] - x[s[u]]));
            else
                t = min(t, abs(y[i] - y[s[u]]));
        result += t * (i64)p[i];
    }
    ans[j] = min(ans[j], result);
}

void sk(int i, int j)
{
    if (i > n)
    {
        ct(j);
        return;
    }
    s[j] = i;
    c[j] = b[i] = 1;
    sk(i + 1, j + 1);
    c[j] = b[i] = 2;
    sk(i + 1, j + 1);
    b[i] = 0;
    sk(i + 1, j);
}

int main()
{
    scanf("%d", &n);
    for (i = 1; i <= n; ++i)
    {
        scanf("%d%d%d", &x[i], &y[i], &p[i]);
        z[i] = min(abs(x[i]), abs(y[i]));
    }
    for (i = 0; i <= n; ++i)
        ans[i] = numeric_limits<i64>::max();
    sk(1, 0);
    for (i = 0; i <= n; ++i)
        printf("%lld\n", ans[i]);
}