#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) FOR(i, 0, (n))
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define LAR(a, b) ((a)=max((a),(b)))
#define SML(a, b) ((a)=min((a),(b)))
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;
template<typename T>
using pque = priority_queue<T, vector<T>, greater<T>>;
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define ALL(a) (a).begin(), (a).end()
#ifdef LOCAL_DEBUG
#define DEBUG(...) printf(__VA_ARGS__)
#else
#define DEBUG(...)
#endif

bool next_one(int *s, int *t, int k){
	while(s < t){
		if(++*s < k) return true;
		*(s++) = 0;
	}
	return false;
}

constexpr int N = 32;
int x[2][N];
ll z[N];

int p[N] = {};
int m[2][1<<15][N];
int q[N] = {};
ll ans[N];

int main(){
	int n; scanf("%d", &n);
	REP(i, n) scanf("%d%d%lld", x[0]+i, x[1]+i, z+i);

	REP(h, 2){
		REP(t, 1<<n){
			REP(i, n){
				int &y = m[h][t][i];
				y = abs(x[h][i]);
				REP(j, n){
					if(t & (1<<j)){
						SML(y, abs(x[h][i]-x[h][j]));
					}
				}
			}
		}
	}

	fill(ans, ans+N, LLONG_MAX);
	do{
		fill(q, q+n, INT_MAX);
		REP(h, 2){
			int t = 0;
			REP(i, n) if(p[i]==h) t|=1<<i;
			REP(i, n) SML(q[i], m[h][t][i]);
		}
		ll sm = 0;
		REP(i, n) sm += z[i] * q[i];
		int b = 0;
		REP(i, n) if(p[i]<2) b++;
		SML(ans[b], sm);
	}while(next_one(p, p+n, 3));

	REP(i, n+1) printf("%lld\n", ans[i]);
}

### Prompt

Please provide a Cpp coded solution 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;
#define REP(i, n) FOR(i, 0, (n))
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define LAR(a, b) ((a)=max((a),(b)))
#define SML(a, b) ((a)=min((a),(b)))
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;
template<typename T>
using pque = priority_queue<T, vector<T>, greater<T>>;
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define ALL(a) (a).begin(), (a).end()
#ifdef LOCAL_DEBUG
#define DEBUG(...) printf(__VA_ARGS__)
#else
#define DEBUG(...)
#endif

bool next_one(int *s, int *t, int k){
	while(s < t){
		if(++*s < k) return true;
		*(s++) = 0;
	}
	return false;
}

constexpr int N = 32;
int x[2][N];
ll z[N];

int p[N] = {};
int m[2][1<<15][N];
int q[N] = {};
ll ans[N];

int main(){
	int n; scanf("%d", &n);
	REP(i, n) scanf("%d%d%lld", x[0]+i, x[1]+i, z+i);

	REP(h, 2){
		REP(t, 1<<n){
			REP(i, n){
				int &y = m[h][t][i];
				y = abs(x[h][i]);
				REP(j, n){
					if(t & (1<<j)){
						SML(y, abs(x[h][i]-x[h][j]));
					}
				}
			}
		}
	}

	fill(ans, ans+N, LLONG_MAX);
	do{
		fill(q, q+n, INT_MAX);
		REP(h, 2){
			int t = 0;
			REP(i, n) if(p[i]==h) t|=1<<i;
			REP(i, n) SML(q[i], m[h][t][i]);
		}
		ll sm = 0;
		REP(i, n) sm += z[i] * q[i];
		int b = 0;
		REP(i, n) if(p[i]<2) b++;
		SML(ans[b], sm);
	}while(next_one(p, p+n, 3));

	REP(i, n+1) printf("%lld\n", ans[i]);
}

```