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

#define for_(i,a,b) for(int i=a;i<b;++i)
#define ALL(a) a.begin(),a.end()
#define INIT(a,b) memset(a,b,sizeof(a))
using lint = long long int;

int N;
lint X[15], Y[15], P[15];
lint dst_x[1 << 15][15], dst_y[1 << 15][15];
lint ans[16];

void rec(int i, int Sx, int Sy, int K) {
    if (i == N) {
        lint val = 0;
        for_(k,0,N) val += min(dst_x[Sx][k], dst_y[Sy][k]) * P[k];
        ans[K] = min(ans[K], val);
        return;
    }

    rec(i + 1, Sx, Sy, K);
    rec(i + 1, Sx | 1 << i, Sy, K + 1);
    rec(i + 1, Sx, Sy | 1 << i, K + 1);
}

int main() {
    cin >> N;
    for_(i,0,N) cin >> X[i] >> Y[i] >> P[i];
    for_(i,0,N+1) ans[i] = 1LL << 60;

    for_(S,0,1<<N) {
        for_(i,0,N) {
            lint mnx = abs(X[i]), mny = abs(Y[i]);
            for_(j,0,N) {
                if (S >> j & 1) {
                    mnx = min(mnx, abs(X[j] - X[i]));
                    mny = min(mny, abs(Y[i] - Y[j]));
                }
            }
            dst_x[S][i] = mnx;
            dst_y[S][i] = mny;
        }
    }

    rec(0, 0, 0, 0);

    for_(i,0,N+1) cout << ans[i] << endl;
}
### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

#define for_(i,a,b) for(int i=a;i<b;++i)
#define ALL(a) a.begin(),a.end()
#define INIT(a,b) memset(a,b,sizeof(a))
using lint = long long int;

int N;
lint X[15], Y[15], P[15];
lint dst_x[1 << 15][15], dst_y[1 << 15][15];
lint ans[16];

void rec(int i, int Sx, int Sy, int K) {
    if (i == N) {
        lint val = 0;
        for_(k,0,N) val += min(dst_x[Sx][k], dst_y[Sy][k]) * P[k];
        ans[K] = min(ans[K], val);
        return;
    }

    rec(i + 1, Sx, Sy, K);
    rec(i + 1, Sx | 1 << i, Sy, K + 1);
    rec(i + 1, Sx, Sy | 1 << i, K + 1);
}

int main() {
    cin >> N;
    for_(i,0,N) cin >> X[i] >> Y[i] >> P[i];
    for_(i,0,N+1) ans[i] = 1LL << 60;

    for_(S,0,1<<N) {
        for_(i,0,N) {
            lint mnx = abs(X[i]), mny = abs(Y[i]);
            for_(j,0,N) {
                if (S >> j & 1) {
                    mnx = min(mnx, abs(X[j] - X[i]));
                    mny = min(mny, abs(Y[i] - Y[j]));
                }
            }
            dst_x[S][i] = mnx;
            dst_y[S][i] = mny;
        }
    }

    rec(0, 0, 0, 0);

    for_(i,0,N+1) cout << ans[i] << endl;
}
```