#include <bits/stdc++.h>
using namespace std;
using Int = long long;
int PREP = (cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(9), 0);
//int SEGV = getenv("D") || (exit(system("D= SEGFAULT_SIGNALS=all catchsegv ./prog.exe") >> 8), 0);
const Int INF = 1e18;
vector<Int> X, Y, P;
Int N, D[30][30], curr, cnt, ans[30];
void recur(int xory, int i) {
    if (i == N && xory == 1) return;
    if (i == N) { recur(1, 0); return; }
    recur(xory, i + 1);
    if (D[i][cnt] != 0) {
        auto& Z = (xory == 0 ? X : Y);
        Int a = 0;
        for (int t = 0; t < N; t++) {
            D[t][cnt + 1] = min(D[t][cnt], abs(Z[i] - Z[t]));
            a += D[t][cnt + 1] * P[t];
        }
        ans[cnt + 1] = min(ans[cnt + 1], a);
        cnt++;
        recur(xory, i + 1);
        cnt--;
    }
}
int main() {
    cin >> N;
    X.resize(N), Y.resize(N), P.resize(N);
    for (int i = 0; i < N; i++) cin >> X[i] >> Y[i] >> P[i];
    for (int i = 0; i < N; i++) {
        D[i][0] = min(abs(X[i]), abs(Y[i]));
        curr += D[i][0] * P[i];
    }
    for (int i = 0; i <= N; i++) {
        ans[i] = INF;
    }
    ans[0] = curr;
    recur(0, 0);
    for (int i = 0; i <= N; i++) {
        cout << (ans[i] == INF ? 0 : ans[i]) << '\n';
    }
    return 0;
}

### Prompt

Create a solution in cpp for 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;
using Int = long long;
int PREP = (cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(9), 0);
//int SEGV = getenv("D") || (exit(system("D= SEGFAULT_SIGNALS=all catchsegv ./prog.exe") >> 8), 0);
const Int INF = 1e18;
vector<Int> X, Y, P;
Int N, D[30][30], curr, cnt, ans[30];
void recur(int xory, int i) {
    if (i == N && xory == 1) return;
    if (i == N) { recur(1, 0); return; }
    recur(xory, i + 1);
    if (D[i][cnt] != 0) {
        auto& Z = (xory == 0 ? X : Y);
        Int a = 0;
        for (int t = 0; t < N; t++) {
            D[t][cnt + 1] = min(D[t][cnt], abs(Z[i] - Z[t]));
            a += D[t][cnt + 1] * P[t];
        }
        ans[cnt + 1] = min(ans[cnt + 1], a);
        cnt++;
        recur(xory, i + 1);
        cnt--;
    }
}
int main() {
    cin >> N;
    X.resize(N), Y.resize(N), P.resize(N);
    for (int i = 0; i < N; i++) cin >> X[i] >> Y[i] >> P[i];
    for (int i = 0; i < N; i++) {
        D[i][0] = min(abs(X[i]), abs(Y[i]));
        curr += D[i][0] * P[i];
    }
    for (int i = 0; i <= N; i++) {
        ans[i] = INF;
    }
    ans[0] = curr;
    recur(0, 0);
    for (int i = 0; i <= N; i++) {
        cout << (ans[i] == INF ? 0 : ans[i]) << '\n';
    }
    return 0;
}

```