#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int N = 2e4 + 11;
const int inf = 1e9;
int head[N], to[N], nex[N], size = 1;
int dl[N], fr[N];
int n, S = 3000, T = 3001;
LL dis[N], wei[N], cst[N], ans;
bool mark[N];
void add(int x, int y, int z, int k){
	to[++size] = y; nex[size] = head[x]; head[x] = size; cst[size] = k; wei[size] = z;
	to[++size] = x; nex[size] = head[y]; head[y] = size; cst[size] = -k; wei[size] = 0;
}
void spfa(){
	int hd = 0, tl = 1;
	dl[1] = S;
	for(int i = 1;i <= T; i++)dis[i] = -1e18;
	dis[S] = 0; mark[S] = 1;
	while(hd < tl){
		int u = dl[++hd];
		for(int i = head[u];i;i = nex[i]){
			if(wei[i] <= 0)continue;
			int v = to[i];
			if(dis[v] < dis[u] + cst[i]){
				dis[v] = dis[u] + cst[i];
				fr[v] = i;
				if(!mark[v])dl[++tl] = v, mark[v] = 1;
			}
		}
		mark[u] = 0;
	}
}
void MCMF(){
	while(1){
		spfa();
		LL ma = dis[T]; 
		if(ma == -1e18)return ;
		LL flow = 1e9;		
//		printf("ma=%lld ans=%lld\n", ma, ans);
		for(int u = T;u != S;u = to[fr[u]^1])flow = min(flow, wei[fr[u]]);
		ans += flow * ma;
		for(int u = T;u != S;u = to[fr[u]^1])wei[fr[u]] -= flow, wei[fr[u]^1] += flow;
	}
}
int main(){
	cin>>n;
	int x, y, w;
	for(int i = 1;i <= n; i++){
		scanf("%d%d%d", &x, &y, &w);
		add(S, i, w, 0);
		add(i, 2 * n + 1, inf, x + y);
		add(i, 2 * n + 2, inf, x - y);
		add(i, 2 * n + 3, inf, -x + y);
		add(i, 2 * n + 4, inf, -x - y);
	}
	for(int i = 1;i <= n; i++){
		scanf("%d%d%d", &x, &y, &w);
		add(i + n, T, w, 0);
		add(2 * n + 1, i + n, inf, -x - y);
		add(2 * n + 2, i + n, inf, -x + y);
		add(2 * n + 3, i + n, inf, x - y);
		add(2 * n + 4, i + n, inf, x + y);
	}
	MCMF();
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Generate a cpp solution to the following problem:

Snuke is playing with red and blue balls, placing them on a two-dimensional plane.

First, he performed N operations to place red balls. In the i-th of these operations, he placed RC_i red balls at coordinates (RX_i,RY_i). Then, he performed another N operations to place blue balls. In the i-th of these operations, he placed BC_i blue balls at coordinates (BX_i,BY_i). The total number of red balls placed and the total number of blue balls placed are equal, that is, \sum_{i=1}^{N} RC_i = \sum_{i=1}^{N} BC_i. Let this value be S.

Snuke will now form S pairs of red and blue balls so that every ball belongs to exactly one pair. Let us define the score of a pair of a red ball at coordinates (rx, ry) and a blue ball at coordinates (bx, by) as |rx-bx| + |ry-by|.

Snuke wants to maximize the sum of the scores of the pairs. Help him by finding the maximum possible sum of the scores of the pairs.

Constraints

* 1 \leq N \leq 1000
* 0 \leq RX_i,RY_i,BX_i,BY_i \leq 10^9
* 1 \leq RC_i,BC_i \leq 10
* \sum_{i=1}^{N} RC_i = \sum_{i=1}^{N} BC_i
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
RX_1 RY_1 RC_1
RX_2 RY_2 RC_2
\vdots
RX_N RY_N RC_N
BX_1 BY_1 BC_1
BX_2 BY_2 BC_2
\vdots
BX_N BY_N BC_N


Output

Print the maximum possible sum of the scores of the pairs.

Examples

Input

2
0 0 1
3 2 1
2 2 1
5 0 1


Output

8


Input

3
0 0 1
2 2 1
0 0 2
1 1 1
1 1 1
3 3 2


Output

16


Input

10
582463373 690528069 8
621230322 318051944 4
356524296 974059503 6
372751381 111542460 9
392867214 581476334 6
606955458 513028121 5
882201596 791660614 9
250465517 91918758 3
618624774 406956634 6
426294747 736401096 5
974896051 888765942 5
726682138 336960821 3
715144179 82444709 6
599055841 501257806 6
390484433 962747856 4
912334580 219343832 8
570458984 648862300 6
638017635 572157978 10
435958984 585073520 7
445612658 234265014 6


Output

45152033546

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int N = 2e4 + 11;
const int inf = 1e9;
int head[N], to[N], nex[N], size = 1;
int dl[N], fr[N];
int n, S = 3000, T = 3001;
LL dis[N], wei[N], cst[N], ans;
bool mark[N];
void add(int x, int y, int z, int k){
	to[++size] = y; nex[size] = head[x]; head[x] = size; cst[size] = k; wei[size] = z;
	to[++size] = x; nex[size] = head[y]; head[y] = size; cst[size] = -k; wei[size] = 0;
}
void spfa(){
	int hd = 0, tl = 1;
	dl[1] = S;
	for(int i = 1;i <= T; i++)dis[i] = -1e18;
	dis[S] = 0; mark[S] = 1;
	while(hd < tl){
		int u = dl[++hd];
		for(int i = head[u];i;i = nex[i]){
			if(wei[i] <= 0)continue;
			int v = to[i];
			if(dis[v] < dis[u] + cst[i]){
				dis[v] = dis[u] + cst[i];
				fr[v] = i;
				if(!mark[v])dl[++tl] = v, mark[v] = 1;
			}
		}
		mark[u] = 0;
	}
}
void MCMF(){
	while(1){
		spfa();
		LL ma = dis[T]; 
		if(ma == -1e18)return ;
		LL flow = 1e9;		
//		printf("ma=%lld ans=%lld\n", ma, ans);
		for(int u = T;u != S;u = to[fr[u]^1])flow = min(flow, wei[fr[u]]);
		ans += flow * ma;
		for(int u = T;u != S;u = to[fr[u]^1])wei[fr[u]] -= flow, wei[fr[u]^1] += flow;
	}
}
int main(){
	cin>>n;
	int x, y, w;
	for(int i = 1;i <= n; i++){
		scanf("%d%d%d", &x, &y, &w);
		add(S, i, w, 0);
		add(i, 2 * n + 1, inf, x + y);
		add(i, 2 * n + 2, inf, x - y);
		add(i, 2 * n + 3, inf, -x + y);
		add(i, 2 * n + 4, inf, -x - y);
	}
	for(int i = 1;i <= n; i++){
		scanf("%d%d%d", &x, &y, &w);
		add(i + n, T, w, 0);
		add(2 * n + 1, i + n, inf, -x - y);
		add(2 * n + 2, i + n, inf, -x + y);
		add(2 * n + 3, i + n, inf, x - y);
		add(2 * n + 4, i + n, inf, x + y);
	}
	MCMF();
	printf("%lld\n", ans);
	return 0;
}
```