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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<ll> x(n), y(n), p(n), s(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> y[i] >> p[i];
        s[i] = min(abs(x[i]), abs(y[i]));
    }
    vector<ll> ans(n + 1, 1e18);
    vector<int> a(n);
    int z = 1;
    for (int i = 0; i < n; i++) {
        z *= 3;
    }
    for (int loop = 0; loop < z; loop++) {
        int b = 0;
        for (int i = 0; i < n; i++) {
            if (a[i]) {
                b++;
            }
        }
        vector<ll> t = s;
        for (int i = 0; i < n; i++) {
            if (a[i] == 1) {
                for (int j = 0; j < n; j++) {
                    t[j] = min(t[j], abs(x[i] - x[j]));
                }
            } else if (a[i] == 2) {
                for (int j = 0; j < n; j++) {
                    t[j] = min(t[j], abs(y[i] - y[j]));
                }
            }
        }
        ll sum = 0;
        for (int i = 0; i < n; i++) {
            sum += t[i] * p[i];
        }
        ans[b] = min(ans[b], sum);
        for (int i = n - 1; i >= 0; i--) {
            a[i]++;
            if (a[i] == 3) {
                a[i] = 0;
            } else {
                break;
            }
        }
    }
    for (ll i : ans) {
        cout << i << '\n';
    }
    return 0;
}