#include <queue>
#include <cstdio>
#include <cstring>

const int maxn = 112345, maxm = 412345;
typedef long long ll;

struct edge{
	int v, nxt, flow;
	ll cost;
};

edge e[maxm]; int fir[maxn], num(1);
int S = 0, T;

inline void add(int u, int v, int flow, ll cost){
	e[++num].v = v; e[num].flow = flow; e[num].cost = cost;
	e[num].nxt = fir[u]; fir[u] = num;
	e[++num].v = u; e[num].flow = 0; e[num].cost = -cost;
	e[num].nxt = fir[v]; fir[v] = num;
}

ll ans(0);

ll dis[maxn]; int pre[maxn], f[maxn];

inline ll max(ll a, ll b){
	return a > b ? a : b;
}

inline ll min(ll a, ll b){
	return a < b ? a : b;
}

bool vis[maxn];

const ll inf = -1e18;

inline bool spfa(){
	std::queue<int> q; q.push(0);
	for(int i(1); i <= T; ++i) dis[i] = inf, f[i] = 0, vis[i] = 0;
	dis[0] = 0; f[0] = 1123456789; vis[0] = 1;
	while(!q.empty()){
		int now = q.front(); q.pop(); vis[now] = 0;
		for(int itr(fir[now]), v; itr; itr = e[itr].nxt){
			if(e[itr].flow == 0) continue;
		 	v = e[itr].v;
		 	/*
			if(dis[now] + e[itr].cost == dis[v] && min(f[now], e[itr].flow) >= f[v]){
				f[v] = min(f[now], e[itr].flow); pre[v] = now; 
				if(!vis[v]){
					vis[v] = true; q.push(v);
				}
			}
			*/
			if(dis[now] + e[itr].cost > dis[v]){
				f[v] = min(f[now], e[itr].flow); pre[v] = itr;
				dis[v] = dis[now] + e[itr].cost;
				if(!vis[v]){
					vis[v] = true; q.push(v);
				}
			}
		}
	}
	return f[T] != 0;
}

inline void process(){
	int now = T;
	while(now != 0){
		int itr = pre[now], oppo = itr^1, nxt = e[oppo].v;
		e[itr].flow -= f[T], e[oppo].flow += f[T];
		now = nxt;
	}
}

inline void EK(){
	while(spfa()){
		process();
		ans += dis[T]*f[T];
	}
}

int main(){
	int n; scanf("%d", &n);
	T = 2*n + 5;
	for(int i(1), t; i <= n; ++i){ ll x, y;
		scanf("%lld%lld%d", &x, &y, &t);
		add(S, i, t, 0);
		add(i, 2*n + 1, t, x + y);
		add(i, 2*n + 2, t, x - y);
		add(i, 2*n + 3, t,-x + y);
		add(i, 2*n + 4, t,-x - y);
	}
	for(int i(1), t; i <= n; ++i){ ll x, y;
		scanf("%lld%lld%d", &x, &y, &t);
		add(i + n, T, t, 0);
		add(2*n + 1, i + n, t,-x - y);
		add(2*n + 2, i + n, t,-x + y);
		add(2*n + 3, i + n, t,x - y);
		add(2*n + 4, i + n, t,x + y);
	}
	EK();
	printf("%lld\n", ans);
	return 0;
}


### Prompt

Your challenge is to write 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 <queue>
#include <cstdio>
#include <cstring>

const int maxn = 112345, maxm = 412345;
typedef long long ll;

struct edge{
	int v, nxt, flow;
	ll cost;
};

edge e[maxm]; int fir[maxn], num(1);
int S = 0, T;

inline void add(int u, int v, int flow, ll cost){
	e[++num].v = v; e[num].flow = flow; e[num].cost = cost;
	e[num].nxt = fir[u]; fir[u] = num;
	e[++num].v = u; e[num].flow = 0; e[num].cost = -cost;
	e[num].nxt = fir[v]; fir[v] = num;
}

ll ans(0);

ll dis[maxn]; int pre[maxn], f[maxn];

inline ll max(ll a, ll b){
	return a > b ? a : b;
}

inline ll min(ll a, ll b){
	return a < b ? a : b;
}

bool vis[maxn];

const ll inf = -1e18;

inline bool spfa(){
	std::queue<int> q; q.push(0);
	for(int i(1); i <= T; ++i) dis[i] = inf, f[i] = 0, vis[i] = 0;
	dis[0] = 0; f[0] = 1123456789; vis[0] = 1;
	while(!q.empty()){
		int now = q.front(); q.pop(); vis[now] = 0;
		for(int itr(fir[now]), v; itr; itr = e[itr].nxt){
			if(e[itr].flow == 0) continue;
		 	v = e[itr].v;
		 	/*
			if(dis[now] + e[itr].cost == dis[v] && min(f[now], e[itr].flow) >= f[v]){
				f[v] = min(f[now], e[itr].flow); pre[v] = now; 
				if(!vis[v]){
					vis[v] = true; q.push(v);
				}
			}
			*/
			if(dis[now] + e[itr].cost > dis[v]){
				f[v] = min(f[now], e[itr].flow); pre[v] = itr;
				dis[v] = dis[now] + e[itr].cost;
				if(!vis[v]){
					vis[v] = true; q.push(v);
				}
			}
		}
	}
	return f[T] != 0;
}

inline void process(){
	int now = T;
	while(now != 0){
		int itr = pre[now], oppo = itr^1, nxt = e[oppo].v;
		e[itr].flow -= f[T], e[oppo].flow += f[T];
		now = nxt;
	}
}

inline void EK(){
	while(spfa()){
		process();
		ans += dis[T]*f[T];
	}
}

int main(){
	int n; scanf("%d", &n);
	T = 2*n + 5;
	for(int i(1), t; i <= n; ++i){ ll x, y;
		scanf("%lld%lld%d", &x, &y, &t);
		add(S, i, t, 0);
		add(i, 2*n + 1, t, x + y);
		add(i, 2*n + 2, t, x - y);
		add(i, 2*n + 3, t,-x + y);
		add(i, 2*n + 4, t,-x - y);
	}
	for(int i(1), t; i <= n; ++i){ ll x, y;
		scanf("%lld%lld%d", &x, &y, &t);
		add(i + n, T, t, 0);
		add(2*n + 1, i + n, t,-x - y);
		add(2*n + 2, i + n, t,-x + y);
		add(2*n + 3, i + n, t,x - y);
		add(2*n + 4, i + n, t,x + y);
	}
	EK();
	printf("%lld\n", ans);
	return 0;
}


```