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

int nth_bit(int64_t num, int n){
    return (num >> n) & 1;
}

void chmin(int64_t& a, int64_t b){
    a = min(a, b);
}

int main(){
    int N;
    cin >> N;
    vector<int64_t> X(N), Y(N), P(N);
    for(int i=0; i<N; i++) cin >> X[i] >> Y[i] >> P[i];

    vector<int64_t> pre(1<<N, 1e18), pre0(1<<N);
    for(int b=0; b<(1<<N); b++){
        vector<pair<int64_t, int64_t>> xp, yp;
        for(int i=0; i<N; i++) if(nth_bit(b, i)){
            xp.emplace_back(X[i], P[i]);
            yp.emplace_back(Y[i], P[i]);
            pre0[b] += min(abs(X[i]), abs(Y[i])) * P[i];
        }
        sort(xp.begin(), xp.end());
        sort(yp.begin(), yp.end());
        for(int i=0; i<N; i++){
            int64_t xs = 0, ys = 0;
            for(auto& p : xp) xs += abs((p.first-X[i])*p.second);
            for(auto& p : yp) ys += abs((p.first-Y[i])*p.second);
            chmin(pre[b], min(xs, ys));
        }
    }

    vector<vector<int64_t>> dp(1<<N, vector<int64_t>(N+1, 1e18));
    for(int b=0; b<(1<<N); b++) for(int k=0; k<=N; k++) dp[b][k] = pre0[b];
    for(int b=0; b<(1<<N); b++) for(int b1=b; b1>0; b1=(b1-1)&b) for(int k=0; k<N; k++) chmin(dp[b][k+1], dp[b-b1][k] + pre[b1]);
    for(int k=0; k<=N; k++) cout << dp[(1<<N)-1][k] << endl;
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;

int nth_bit(int64_t num, int n){
    return (num >> n) & 1;
}

void chmin(int64_t& a, int64_t b){
    a = min(a, b);
}

int main(){
    int N;
    cin >> N;
    vector<int64_t> X(N), Y(N), P(N);
    for(int i=0; i<N; i++) cin >> X[i] >> Y[i] >> P[i];

    vector<int64_t> pre(1<<N, 1e18), pre0(1<<N);
    for(int b=0; b<(1<<N); b++){
        vector<pair<int64_t, int64_t>> xp, yp;
        for(int i=0; i<N; i++) if(nth_bit(b, i)){
            xp.emplace_back(X[i], P[i]);
            yp.emplace_back(Y[i], P[i]);
            pre0[b] += min(abs(X[i]), abs(Y[i])) * P[i];
        }
        sort(xp.begin(), xp.end());
        sort(yp.begin(), yp.end());
        for(int i=0; i<N; i++){
            int64_t xs = 0, ys = 0;
            for(auto& p : xp) xs += abs((p.first-X[i])*p.second);
            for(auto& p : yp) ys += abs((p.first-Y[i])*p.second);
            chmin(pre[b], min(xs, ys));
        }
    }

    vector<vector<int64_t>> dp(1<<N, vector<int64_t>(N+1, 1e18));
    for(int b=0; b<(1<<N); b++) for(int k=0; k<=N; k++) dp[b][k] = pre0[b];
    for(int b=0; b<(1<<N); b++) for(int b1=b; b1>0; b1=(b1-1)&b) for(int k=0; k<N; k++) chmin(dp[b][k+1], dp[b-b1][k] + pre[b1]);
    for(int k=0; k<=N; k++) cout << dp[(1<<N)-1][k] << endl;
    return 0;
}

```