#include <bits/stdc++.h>

#define int LL

typedef long long LL;
const int N = 2010, INF = 0x3f3f3f3f3f3f3f3fll;

struct Edge {
	int nex, v, c, len;
	Edge(){}
	Edge(int N, int V, int C, int L) {
		nex = N, v = V, c = C, len = L;
	}
}edge[2000010]; int tp = 1;

int e[N], n, tot, d[N], pre[N], flow[N], Time, vis[N];
std::queue<int> Q;

inline void add(int x, int y, int z, int w) {
	edge[++tp] = Edge(e[x], y, z, w);
	e[x] = tp;
	edge[++tp] = Edge(e[y], x, 0, -w);
	e[y] = tp;
	return;
}

inline bool SPFA(int s, int t) {
	memset(d, 0x3f, sizeof(d));
	Q.push(s);
	++Time;
	d[s] = 0;
	vis[s] = Time;
	flow[s] = INF;
	while(Q.size()) {
		int x = Q.front();
		Q.pop();
		vis[x] = 0;
		for(int i = e[x]; i; i = edge[i].nex) {
			int y = edge[i].v;
			if(edge[i].c && d[y] > d[x] + edge[i].len) {
				d[y] = d[x] + edge[i].len;
				pre[y] = i;
				flow[y] = std::min(flow[x], edge[i].c);
				if(vis[y] != Time) {
					vis[y] = Time;
					Q.push(y);
				}
			}
		}
	}
	return d[t] < INF;
}

inline void update(int s, int t) {
	int f = flow[t];
	while(s != t) {
		int i = pre[t];
		edge[i].c -= f;
		edge[i ^ 1].c += f;
		t = edge[i ^ 1].v;
	}
	return;
}

inline int solve(int s, int t, int &cost) {
	cost = 0;
	int ans = 0;
	while(SPFA(s, t)) {
		//printf("!");
		ans += flow[t];
		cost += flow[t] * d[t];
		update(s, t);
	}
	return ans;
}

signed main() {
	
	scanf("%lld", &n);
	int s = 2 * n + 5, t = s + 1, x, y, z;
	for(int i = 1; i <= n; i++) {
		scanf("%lld%lld%lld", &x, &y, &z);
		add(s, i, z, 0);
		add(i, 2 * n + 1, z, x + y);
		add(i, 2 * n + 2, z, y - x);
		add(i, 2 * n + 3, z, x - y);
		add(i, 2 * n + 4, z, -x - y);
	}
	for(int i = 1; i <= n; i++) {
		scanf("%lld%lld%lld", &x, &y, &z);
		add(n + i, t, z, 0);
		add(2 * n + 1, n + i, z, -x - y);
		add(2 * n + 2, n + i, z, x - y);
		add(2 * n + 3, n + i, z, y - x);
		add(2 * n + 4, n + i, z, x + y);
	}
	//puts("OVER");
	int cost = 0;
	solve(s, t, cost);
	printf("%lld\n", -cost);
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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>

#define int LL

typedef long long LL;
const int N = 2010, INF = 0x3f3f3f3f3f3f3f3fll;

struct Edge {
	int nex, v, c, len;
	Edge(){}
	Edge(int N, int V, int C, int L) {
		nex = N, v = V, c = C, len = L;
	}
}edge[2000010]; int tp = 1;

int e[N], n, tot, d[N], pre[N], flow[N], Time, vis[N];
std::queue<int> Q;

inline void add(int x, int y, int z, int w) {
	edge[++tp] = Edge(e[x], y, z, w);
	e[x] = tp;
	edge[++tp] = Edge(e[y], x, 0, -w);
	e[y] = tp;
	return;
}

inline bool SPFA(int s, int t) {
	memset(d, 0x3f, sizeof(d));
	Q.push(s);
	++Time;
	d[s] = 0;
	vis[s] = Time;
	flow[s] = INF;
	while(Q.size()) {
		int x = Q.front();
		Q.pop();
		vis[x] = 0;
		for(int i = e[x]; i; i = edge[i].nex) {
			int y = edge[i].v;
			if(edge[i].c && d[y] > d[x] + edge[i].len) {
				d[y] = d[x] + edge[i].len;
				pre[y] = i;
				flow[y] = std::min(flow[x], edge[i].c);
				if(vis[y] != Time) {
					vis[y] = Time;
					Q.push(y);
				}
			}
		}
	}
	return d[t] < INF;
}

inline void update(int s, int t) {
	int f = flow[t];
	while(s != t) {
		int i = pre[t];
		edge[i].c -= f;
		edge[i ^ 1].c += f;
		t = edge[i ^ 1].v;
	}
	return;
}

inline int solve(int s, int t, int &cost) {
	cost = 0;
	int ans = 0;
	while(SPFA(s, t)) {
		//printf("!");
		ans += flow[t];
		cost += flow[t] * d[t];
		update(s, t);
	}
	return ans;
}

signed main() {
	
	scanf("%lld", &n);
	int s = 2 * n + 5, t = s + 1, x, y, z;
	for(int i = 1; i <= n; i++) {
		scanf("%lld%lld%lld", &x, &y, &z);
		add(s, i, z, 0);
		add(i, 2 * n + 1, z, x + y);
		add(i, 2 * n + 2, z, y - x);
		add(i, 2 * n + 3, z, x - y);
		add(i, 2 * n + 4, z, -x - y);
	}
	for(int i = 1; i <= n; i++) {
		scanf("%lld%lld%lld", &x, &y, &z);
		add(n + i, t, z, 0);
		add(2 * n + 1, n + i, z, -x - y);
		add(2 * n + 2, n + i, z, x - y);
		add(2 * n + 3, n + i, z, y - x);
		add(2 * n + 4, n + i, z, x + y);
	}
	//puts("OVER");
	int cost = 0;
	solve(s, t, cost);
	printf("%lld\n", -cost);
	return 0;
}
```