#include <bits/stdc++.h>
using namespace std;

const int MAXN = 20;

int n, x[MAXN], y[MAXN];
long long p[MAXN], dp1[1 << 15], dp[MAXN][1 << 15];

int main(){
	scanf("%d", &n);
	for(int i = 0; i < n; i++)
		scanf("%d %d %lld", &x[i], &y[i], &p[i]);
	
	for(int mask = 0; mask < (1 << n); mask++){
		if(mask == 0)dp1[mask] = 0;
		else{
			dp1[mask] = 1LL << 50;
			for(int i = 0; i < n; i++)if(mask & (1 << i)){
				long long r0 = 0, r1 = 0;
				for(int j = 0; j < n; j++)if(mask & (1 << j)){
					int d = min(abs(x[j]), abs(y[j]));
					r0 += p[j] * min(d, abs(x[j] - x[i]));
					r1 += p[j] * min(d, abs(y[j] - y[i]));
				}
				dp1[mask] = min(dp1[mask], min(r0, r1));
			}
		}
	}
	
	for(int k = 0; k <= n; k++)
	for(int mask = 0; mask < (1 << n); mask++){
		dp[k][mask] = 0;
		for(int i = 0; i < n; i++)if(mask & (1 << i)){
			dp[k][mask] += p[i] * min(abs(x[i]), abs(y[i]));
		}
		
		if(k > 0){
			for(int sub = mask; sub > 0; sub = (sub - 1) & mask){
				dp[k][mask] = min(dp[k][mask], dp[k - 1][mask ^ sub] + dp1[sub]);
			}
		}
	}
	
	for(int k = 0; k <= n; k++){
		printf("%lld\n", dp[k][(1 << n) - 1]);
	}
	
	return 0;
}

### 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>
using namespace std;

const int MAXN = 20;

int n, x[MAXN], y[MAXN];
long long p[MAXN], dp1[1 << 15], dp[MAXN][1 << 15];

int main(){
	scanf("%d", &n);
	for(int i = 0; i < n; i++)
		scanf("%d %d %lld", &x[i], &y[i], &p[i]);
	
	for(int mask = 0; mask < (1 << n); mask++){
		if(mask == 0)dp1[mask] = 0;
		else{
			dp1[mask] = 1LL << 50;
			for(int i = 0; i < n; i++)if(mask & (1 << i)){
				long long r0 = 0, r1 = 0;
				for(int j = 0; j < n; j++)if(mask & (1 << j)){
					int d = min(abs(x[j]), abs(y[j]));
					r0 += p[j] * min(d, abs(x[j] - x[i]));
					r1 += p[j] * min(d, abs(y[j] - y[i]));
				}
				dp1[mask] = min(dp1[mask], min(r0, r1));
			}
		}
	}
	
	for(int k = 0; k <= n; k++)
	for(int mask = 0; mask < (1 << n); mask++){
		dp[k][mask] = 0;
		for(int i = 0; i < n; i++)if(mask & (1 << i)){
			dp[k][mask] += p[i] * min(abs(x[i]), abs(y[i]));
		}
		
		if(k > 0){
			for(int sub = mask; sub > 0; sub = (sub - 1) & mask){
				dp[k][mask] = min(dp[k][mask], dp[k - 1][mask ^ sub] + dp1[sub]);
			}
		}
	}
	
	for(int k = 0; k <= n; k++){
		printf("%lld\n", dp[k][(1 << n) - 1]);
	}
	
	return 0;
}

```