#include <cstdio>
#include <algorithm>
#define N 100010
#define inf 2000000000
#define ll long long
#define mem(x, a) memset(x, a, sizeof x)
#define fo(x, a, b) for (int x = (a); x <= (b); x++)
#define fd(x, a, b) for (int x = (a); x >= (b); x--)
#define go(x) for (int p = tail[x], v; p; p = e[p].fr)
using namespace std;
struct node{int v, fr, flow, c;}e[N << 3];
const int zf[4][2] = {{1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int n, S, T, cp[4], tail[N << 2], cnt = 1, f[N << 4], tot_node;
bool tag[N << 2], vis[N << 2];
ll dis[N << 2], ans = 0;

inline int read() {
	int x = 0, f = 0; char c = getchar();
	while (c < '0' || c > '9') f = (c == '-') ? 1 : f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
	return f ? -x : x;
}

inline void add(int u, int v, int flow, int c) {
//	printf("%d %d %d\n", u, v, c);
	e[++cnt] = (node){v, tail[u], flow, c}, tail[u] = cnt;
	e[++cnt] = (node){u, tail[v], 0, -c}, tail[v] = cnt;
}

bool spfa() {
	fo(i, 1, tot_node) dis[i] = -inf;
	int l = 0, r = 1; f[1] = S;
	dis[S] = 0, tag[S] = 1;
	while (l++ < r) {
		int x = f[l];
		go(x) {
			if (! e[p].flow) continue;
			if (dis[x] + e[p].c > dis[v = e[p].v]) {
				dis[v] = dis[x] + e[p].c;
				if (! tag[v]) f[++r] = v, tag[v] = 1;
			}
		}
		tag[x] = 0;
	}
	return dis[T] > 0;
}

int dfs(int x, int hav) {
	if (x == T) return hav;
	int gone = 0, can = 0; vis[x] = 1;
	go(x) {
		v = e[p].v;
		if (dis[x] + e[p].c != dis[v] || ! e[p].flow || vis[v]) continue;
		can = dfs(v, min(e[p].flow, hav - gone));
		if (! can) {dis[v] = -inf; continue;}
		gone += can, e[p].flow -= can, e[p ^ 1].flow += can;
		if (gone == hav) {vis[x] = 0; return gone;}
	}
	vis[x] = 0; return gone;
}

int main()
{
	n = read();
	fo(i, 0, 3) cp[i] = (n << 1) + i + 1;
	S = (n << 1) + 5, tot_node = T = (n << 1) + 6;
	fo(i, 1, n) {
		int x = read(), y = read(), c = read(); add(S, i, c, 0);
		fo(j, 0, 3) add(i, cp[j], c, zf[j][0] * x + zf[j][1] * y);
	}
	fo(i, n + 1, (n << 1)) {
		int x = read(), y = read(), c = read(); add(i, T, c, 0);
		fo(j, 0, 3) add(cp[j], i, c, -zf[j][0] * x + -zf[j][1] * y);
	}
	while (spfa()) {
		ll flow = dfs(S, inf);
		ans += flow * dis[T];
	}
	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 <cstdio>
#include <algorithm>
#define N 100010
#define inf 2000000000
#define ll long long
#define mem(x, a) memset(x, a, sizeof x)
#define fo(x, a, b) for (int x = (a); x <= (b); x++)
#define fd(x, a, b) for (int x = (a); x >= (b); x--)
#define go(x) for (int p = tail[x], v; p; p = e[p].fr)
using namespace std;
struct node{int v, fr, flow, c;}e[N << 3];
const int zf[4][2] = {{1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int n, S, T, cp[4], tail[N << 2], cnt = 1, f[N << 4], tot_node;
bool tag[N << 2], vis[N << 2];
ll dis[N << 2], ans = 0;

inline int read() {
	int x = 0, f = 0; char c = getchar();
	while (c < '0' || c > '9') f = (c == '-') ? 1 : f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
	return f ? -x : x;
}

inline void add(int u, int v, int flow, int c) {
//	printf("%d %d %d\n", u, v, c);
	e[++cnt] = (node){v, tail[u], flow, c}, tail[u] = cnt;
	e[++cnt] = (node){u, tail[v], 0, -c}, tail[v] = cnt;
}

bool spfa() {
	fo(i, 1, tot_node) dis[i] = -inf;
	int l = 0, r = 1; f[1] = S;
	dis[S] = 0, tag[S] = 1;
	while (l++ < r) {
		int x = f[l];
		go(x) {
			if (! e[p].flow) continue;
			if (dis[x] + e[p].c > dis[v = e[p].v]) {
				dis[v] = dis[x] + e[p].c;
				if (! tag[v]) f[++r] = v, tag[v] = 1;
			}
		}
		tag[x] = 0;
	}
	return dis[T] > 0;
}

int dfs(int x, int hav) {
	if (x == T) return hav;
	int gone = 0, can = 0; vis[x] = 1;
	go(x) {
		v = e[p].v;
		if (dis[x] + e[p].c != dis[v] || ! e[p].flow || vis[v]) continue;
		can = dfs(v, min(e[p].flow, hav - gone));
		if (! can) {dis[v] = -inf; continue;}
		gone += can, e[p].flow -= can, e[p ^ 1].flow += can;
		if (gone == hav) {vis[x] = 0; return gone;}
	}
	vis[x] = 0; return gone;
}

int main()
{
	n = read();
	fo(i, 0, 3) cp[i] = (n << 1) + i + 1;
	S = (n << 1) + 5, tot_node = T = (n << 1) + 6;
	fo(i, 1, n) {
		int x = read(), y = read(), c = read(); add(S, i, c, 0);
		fo(j, 0, 3) add(i, cp[j], c, zf[j][0] * x + zf[j][1] * y);
	}
	fo(i, n + 1, (n << 1)) {
		int x = read(), y = read(), c = read(); add(i, T, c, 0);
		fo(j, 0, 3) add(cp[j], i, c, -zf[j][0] * x + -zf[j][1] * y);
	}
	while (spfa()) {
		ll flow = dfs(S, inf);
		ans += flow * dis[T];
	}
	printf("%lld\n", ans);
	return 0;
}
```