#include <bits/stdc++.h>

using std::cerr;
using std::endl;

const int N = 2010, INF = 1e9;

int n, S, T, a, b, c, d;

struct Graph {
  int to, nxt, flow, cost;
} G[N * 16];

int head[N * 8], cur[N * 8], tot = 1;

inline void addedge(int x, int y, int f, int c) {
  G[++tot] = { y, head[x], f, -c }, head[x] = tot;
  G[++tot] = { x, head[y], 0, c }, head[y] = tot;
}

bool vis[N * 8];
long long dis[N * 8];
int que[N * 8], ql, qr;

inline bool spfa() {
  memset(vis, 0, sizeof(bool) * (T + 1));
  memset(dis, 0x3f, sizeof(long long) * (T + 1));
  for (int i = 1; i <= T; ++i)
    cur[i] = head[i];
  dis[S] = 0, que[ql = qr = 0] = S;
  while (ql <= qr) {
    int x = que[ql++];
    vis[x] = 0;
    for (int i = head[x]; i; i = G[i].nxt) {
      int y = G[i].to;
      if (G[i].flow && dis[y] > dis[x] + G[i].cost) {
        dis[y] = dis[x] + G[i].cost;
        if (!vis[y]) {
          vis[y] = 1;
          que[++qr] = y;
        }
      }
    }
  }
  return dis[T] != dis[0];
}

int dfs(int x, int mf) {
  if (x == T) return mf;
  int used = 0;
  vis[x] = 1;
  for (int &i = cur[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (G[i].flow && !vis[y] && dis[y] == dis[x] + G[i].cost) {
      int tmp = dfs(y, std::min(G[i].flow, mf - used));
      if (tmp) {
        G[i].flow -= tmp;
        G[i ^ 1].flow += tmp;
        if ((used += tmp) == mf)
          break;
      }
    }
  }
  vis[x] = 0;
  return used;
}

inline long long mcmf() {
  long long ret = 0;
  int flow = 0;
  while (spfa()) {
    int tmp = dfs(S, INF);
    flow += tmp;
    ret += tmp * dis[T];
  }
  cerr << flow << endl;
  return ret;
}

int main() {
  // freopen("data", "r", stdin);
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cin >> n;
  a = n * 2 + 1, b = n * 2 + 2,
  c = n * 2 + 3, d = n * 2 + 4,
  S = n * 2 + 5, T = n * 2 + 6;
  for (int i = 1, x, y, w; i <= n; ++i) {
    std::cin >> x >> y >> w;
    addedge(S, i, w, 0);
    addedge(i, a, w, x + y);
    addedge(i, b, w, x - y);
    addedge(i, c, w, y - x);
    addedge(i, d, w, -y - x);
  }
  for (int i = 1, x, y, w; i <= n; ++i) {
    std::cin >> x >> y >> w;
    addedge(n + i, T, w, 0);
    addedge(a, n + i, w, -y - x);
    addedge(b, n + i, w, -x + y);
    addedge(c, n + i, w, x - y);
    addedge(d, n + i, w, x + y);
  }
  std::cout << -mcmf() << '\n';
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>

using std::cerr;
using std::endl;

const int N = 2010, INF = 1e9;

int n, S, T, a, b, c, d;

struct Graph {
  int to, nxt, flow, cost;
} G[N * 16];

int head[N * 8], cur[N * 8], tot = 1;

inline void addedge(int x, int y, int f, int c) {
  G[++tot] = { y, head[x], f, -c }, head[x] = tot;
  G[++tot] = { x, head[y], 0, c }, head[y] = tot;
}

bool vis[N * 8];
long long dis[N * 8];
int que[N * 8], ql, qr;

inline bool spfa() {
  memset(vis, 0, sizeof(bool) * (T + 1));
  memset(dis, 0x3f, sizeof(long long) * (T + 1));
  for (int i = 1; i <= T; ++i)
    cur[i] = head[i];
  dis[S] = 0, que[ql = qr = 0] = S;
  while (ql <= qr) {
    int x = que[ql++];
    vis[x] = 0;
    for (int i = head[x]; i; i = G[i].nxt) {
      int y = G[i].to;
      if (G[i].flow && dis[y] > dis[x] + G[i].cost) {
        dis[y] = dis[x] + G[i].cost;
        if (!vis[y]) {
          vis[y] = 1;
          que[++qr] = y;
        }
      }
    }
  }
  return dis[T] != dis[0];
}

int dfs(int x, int mf) {
  if (x == T) return mf;
  int used = 0;
  vis[x] = 1;
  for (int &i = cur[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (G[i].flow && !vis[y] && dis[y] == dis[x] + G[i].cost) {
      int tmp = dfs(y, std::min(G[i].flow, mf - used));
      if (tmp) {
        G[i].flow -= tmp;
        G[i ^ 1].flow += tmp;
        if ((used += tmp) == mf)
          break;
      }
    }
  }
  vis[x] = 0;
  return used;
}

inline long long mcmf() {
  long long ret = 0;
  int flow = 0;
  while (spfa()) {
    int tmp = dfs(S, INF);
    flow += tmp;
    ret += tmp * dis[T];
  }
  cerr << flow << endl;
  return ret;
}

int main() {
  // freopen("data", "r", stdin);
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cin >> n;
  a = n * 2 + 1, b = n * 2 + 2,
  c = n * 2 + 3, d = n * 2 + 4,
  S = n * 2 + 5, T = n * 2 + 6;
  for (int i = 1, x, y, w; i <= n; ++i) {
    std::cin >> x >> y >> w;
    addedge(S, i, w, 0);
    addedge(i, a, w, x + y);
    addedge(i, b, w, x - y);
    addedge(i, c, w, y - x);
    addedge(i, d, w, -y - x);
  }
  for (int i = 1, x, y, w; i <= n; ++i) {
    std::cin >> x >> y >> w;
    addedge(n + i, T, w, 0);
    addedge(a, n + i, w, -y - x);
    addedge(b, n + i, w, -x + y);
    addedge(c, n + i, w, x - y);
    addedge(d, n + i, w, x + y);
  }
  std::cout << -mcmf() << '\n';
  return 0;
}

```