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:
#pragma G++ optimize(O2)
#pragma G++ optimize(O3)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 16;
LL res[N];
struct node {
    int x, y, p;
}a[N];
int vis[N];
int dis[N];

void solve() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].p);
    }
    memset(res, 0x3f, sizeof res);

    // i：选择哪几个点上添加横线，1为添加，0为不添加
    for(int i = 0; i < (1 << n); i++) {
        int cnt = 0;
        for(int j = 0; j < n; j++) if((i >> j) & 1) vis[++cnt] = j + 1;

        // j:选择的点的所有状态，1为x方向，0为y方向
        for(int j = 0; j < (1 << cnt); j++) {

            for(int k = 1; k <= n; k++) dis[k] = min(abs(a[k].x), abs(a[k].y));

            for(int k = 0; k < cnt; k++) {
                if((j >> k) & 1) {
                    for(int t = 1; t <= n; t++) {
                        dis[t] = min(dis[t], abs(a[t].x - a[vis[k + 1]].x));
                    }
                }
                else {
                    for(int t = 1; t <= n; t++) {
                        dis[t] = min(dis[t], abs(a[t].y - a[vis[k + 1]].y));
                    }
                }
            }
            LL sum = 0;
            for(int k = 1; k <= n; k++) sum += (LL)dis[k] * a[k].p;
            res[cnt] = min(res[cnt], sum);
        }
    }
    for(int i = 0; i <= n; i++) printf("%lld\n", res[i]);
}

int main() {
//    freopen("in.txt", "r", stdin);
    solve();
    return 0;
}