#include<bits/stdc++.h>
using namespace std;
const int maxn = 15 + 5;
#define ll long long
struct Node
{
    ll x , y;
    ll p;
}node[maxn];
ll minn[maxn];
int n ;
ll ans = 1e18;
void dfs (int step , int rest , ll sum)
{
    if (step == n + 1){
    //    cout << sum << endl;
        ans = min (ans , sum);
        return ;
    }
    // 不放
    dfs (step + 1 , rest , sum);
    if (rest == 0) return ;
    // 横着放 - 更新minn数组
    ll all = 0 , o[16];
    for (int i = 1 ; i <= n ; i++){
        o[i] = minn[i];
        minn[i] = min (minn[i] , abs(node[step].y - node[i].y) * node[i].p);
        all += minn[i];
    }
    dfs (step + 1 , rest - 1 , all);
    // 竖着放
    all = 0;
    for (int i = 1 ; i <= n ; i++){
        minn[i] = o[i];
        minn[i] = min (minn[i] , abs(node[step].x - node[i].x) * node[i].p);
        all += minn[i];
    }
    dfs(step + 1 , rest - 1 , all);
    for (int i = 1 ; i <= n ; i++) minn[i] = o[i];
    return ;
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    memset( minn , 0x3f , sizeof minn);
    ll sum = 0;
    for (int i = 1 ; i <= n ; i++){
        cin >> node[i].x >> node[i].y >> node[i].p;
        minn[i] = min(abs(node[i].x) , abs(node[i].y)) * node[i].p;
        sum += minn[i];
    }
    for (int i = 0 ; i <= n ; i++){
        ans = 1e18;
        dfs(1 , i , sum);
        cout << ans << 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;
const int maxn = 15 + 5;
#define ll long long
struct Node
{
    ll x , y;
    ll p;
}node[maxn];
ll minn[maxn];
int n ;
ll ans = 1e18;
void dfs (int step , int rest , ll sum)
{
    if (step == n + 1){
    //    cout << sum << endl;
        ans = min (ans , sum);
        return ;
    }
    // 不放
    dfs (step + 1 , rest , sum);
    if (rest == 0) return ;
    // 横着放 - 更新minn数组
    ll all = 0 , o[16];
    for (int i = 1 ; i <= n ; i++){
        o[i] = minn[i];
        minn[i] = min (minn[i] , abs(node[step].y - node[i].y) * node[i].p);
        all += minn[i];
    }
    dfs (step + 1 , rest - 1 , all);
    // 竖着放
    all = 0;
    for (int i = 1 ; i <= n ; i++){
        minn[i] = o[i];
        minn[i] = min (minn[i] , abs(node[step].x - node[i].x) * node[i].p);
        all += minn[i];
    }
    dfs(step + 1 , rest - 1 , all);
    for (int i = 1 ; i <= n ; i++) minn[i] = o[i];
    return ;
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    memset( minn , 0x3f , sizeof minn);
    ll sum = 0;
    for (int i = 1 ; i <= n ; i++){
        cin >> node[i].x >> node[i].y >> node[i].p;
        minn[i] = min(abs(node[i].x) , abs(node[i].y)) * node[i].p;
        sum += minn[i];
    }
    for (int i = 0 ; i <= n ; i++){
        ans = 1e18;
        dfs(1 , i , sum);
        cout << ans << endl;
    }
    return 0;
}

```