#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

int x[20], y[20], p[20];
ll ans[20], c[(1<<15)+10][18][2];

int main() {
    ios_base::sync_with_stdio(0);
    int n; cin >> n;
    for (int i=0;i<n;i++) cin >> x[i] >> y[i] >> p[i];

    int cnt2 = 1, cnt3 = 1;
    for (int i=0;i<n;i++) {
        cnt2 *= 2; cnt3 *= 3;
    }

    for (int i=0;i<cnt2;i++) {
        vector<int> t;
        for (int j=0;j<n;j++) {
            if (((1<<j)&i)==0) continue;
            t.pb(j);
        }

        // y line
        for (int j=0;j<n;j++) {
            ll now = 1LL*p[j]*abs(x[j]);
            for (auto g: t) now = min(now, 1LL*p[j]*abs(x[g]-x[j]));
            c[i][j][0] = now;
        }

        // x line
        for (int j=0;j<n;j++) {
            ll now = 1LL*p[j]*abs(y[j]);
            for (auto g: t) now = min(now, 1LL*p[j]*abs(y[g]-y[j]));
            c[i][j][1] = now;
        }
    }

    for (int i=0;i<=n;i++) ans[i] = 1e18;

    for (int i=0;i<cnt3;i++) {
        string t = "";
        int v = i, cnt = 0;
        for (int j=0;j<n;j++) {
            t += (char)(v%3+'0');
            if (v%3) cnt++;
            v /= 3;
        }
        reverse(t.begin(), t.end());

        int g1 = 0, g2 = 0, p = 1;
        for (int j=0;j<n;j++) {
            if (t[j]=='1') g1 += p;
            if (t[j]=='2') g2 += p;
            p <<= 1;
        }

        ll now = 0;
        for (int j=0;j<n;j++) {
            now += min(c[g1][j][0], c[g2][j][1]);
        }
        ans[cnt] = min(ans[cnt], now);
    }

    for (int i=0;i<=n;i++) cout << ans[i] << endl;
}

### Prompt

Please formulate 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>
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

int x[20], y[20], p[20];
ll ans[20], c[(1<<15)+10][18][2];

int main() {
    ios_base::sync_with_stdio(0);
    int n; cin >> n;
    for (int i=0;i<n;i++) cin >> x[i] >> y[i] >> p[i];

    int cnt2 = 1, cnt3 = 1;
    for (int i=0;i<n;i++) {
        cnt2 *= 2; cnt3 *= 3;
    }

    for (int i=0;i<cnt2;i++) {
        vector<int> t;
        for (int j=0;j<n;j++) {
            if (((1<<j)&i)==0) continue;
            t.pb(j);
        }

        // y line
        for (int j=0;j<n;j++) {
            ll now = 1LL*p[j]*abs(x[j]);
            for (auto g: t) now = min(now, 1LL*p[j]*abs(x[g]-x[j]));
            c[i][j][0] = now;
        }

        // x line
        for (int j=0;j<n;j++) {
            ll now = 1LL*p[j]*abs(y[j]);
            for (auto g: t) now = min(now, 1LL*p[j]*abs(y[g]-y[j]));
            c[i][j][1] = now;
        }
    }

    for (int i=0;i<=n;i++) ans[i] = 1e18;

    for (int i=0;i<cnt3;i++) {
        string t = "";
        int v = i, cnt = 0;
        for (int j=0;j<n;j++) {
            t += (char)(v%3+'0');
            if (v%3) cnt++;
            v /= 3;
        }
        reverse(t.begin(), t.end());

        int g1 = 0, g2 = 0, p = 1;
        for (int j=0;j<n;j++) {
            if (t[j]=='1') g1 += p;
            if (t[j]=='2') g2 += p;
            p <<= 1;
        }

        ll now = 0;
        for (int j=0;j<n;j++) {
            now += min(c[g1][j][0], c[g2][j][1]);
        }
        ans[cnt] = min(ans[cnt], now);
    }

    for (int i=0;i<=n;i++) cout << ans[i] << endl;
}

```