#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using P = pair<int, int>;
using PL = pair<lint, lint>;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()
constexpr int MOD = 1000000007;
constexpr int INF = 2147483647;
void yes(bool expr) {cout << (expr ? "Yes" : "No") << "\n";}
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  lint N;
  cin >> N;
  vector<lint> X(N);
  vector<lint> Y(N);
  vector<lint> P(N);
  REP(i, N) cin >> X[i] >> Y[i] >> P[i];
  vector<vector<lint>> xd(N, vector<lint>(1<<N));
  REP(i, N) REP(bit, 1<<N) {
    xd[i][bit] = abs(X[i]) * P[i];
    REP(j, N) if(bit>>j&1) xd[i][bit] = min(xd[i][bit], abs(X[i] - X[j]) * P[i]);
  }
  vector<vector<lint>> yd(N, vector<lint>(1<<N));
  REP(i, N) REP(bit, 1<<N) {
    yd[i][bit] = abs(Y[i]) * P[i];
    REP(j, N) if(bit>>j&1) yd[i][bit] = min(yd[i][bit], abs(Y[i] - Y[j]) * P[i]);
  }
  vector<lint> ans(N+1, 1e18);
  vector<lint> pow3(N+1);
  REP(i, N+1) pow3[i] = pow(3LL, i);
  REP(bit, pow3[N]) {
    lint cnt = 0;
    lint xbit = 0;
    lint ybit = 0;
    lint bittmp = bit;
    REP(i, N) {
      lint t = bittmp % 3;
      if(t == 1) {
        cnt++;
        xbit += 1<<i;
      } else if(t == 2) {
        cnt++;
        ybit += 1<<i;
      }
      bittmp /= 3;
    }
    lint sum = 0;
    REP(i, N) sum += min(xd[i][xbit], yd[i][ybit]);
    ans[cnt] = min(ans[cnt], sum);
  }
  REP(i, N+1) cout << ans[i] << "\n";
}
### Prompt

Your challenge is to write 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;
using lint = long long int;
using P = pair<int, int>;
using PL = pair<lint, lint>;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()
constexpr int MOD = 1000000007;
constexpr int INF = 2147483647;
void yes(bool expr) {cout << (expr ? "Yes" : "No") << "\n";}
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  lint N;
  cin >> N;
  vector<lint> X(N);
  vector<lint> Y(N);
  vector<lint> P(N);
  REP(i, N) cin >> X[i] >> Y[i] >> P[i];
  vector<vector<lint>> xd(N, vector<lint>(1<<N));
  REP(i, N) REP(bit, 1<<N) {
    xd[i][bit] = abs(X[i]) * P[i];
    REP(j, N) if(bit>>j&1) xd[i][bit] = min(xd[i][bit], abs(X[i] - X[j]) * P[i]);
  }
  vector<vector<lint>> yd(N, vector<lint>(1<<N));
  REP(i, N) REP(bit, 1<<N) {
    yd[i][bit] = abs(Y[i]) * P[i];
    REP(j, N) if(bit>>j&1) yd[i][bit] = min(yd[i][bit], abs(Y[i] - Y[j]) * P[i]);
  }
  vector<lint> ans(N+1, 1e18);
  vector<lint> pow3(N+1);
  REP(i, N+1) pow3[i] = pow(3LL, i);
  REP(bit, pow3[N]) {
    lint cnt = 0;
    lint xbit = 0;
    lint ybit = 0;
    lint bittmp = bit;
    REP(i, N) {
      lint t = bittmp % 3;
      if(t == 1) {
        cnt++;
        xbit += 1<<i;
      } else if(t == 2) {
        cnt++;
        ybit += 1<<i;
      }
      bittmp /= 3;
    }
    lint sum = 0;
    REP(i, N) sum += min(xd[i][xbit], yd[i][ybit]);
    ans[cnt] = min(ans[cnt], sum);
  }
  REP(i, N+1) cout << ans[i] << "\n";
}
```