// Written by newbiechd
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#define int long long
using namespace std;

const int BUF = 1000003;
inline char getChar() {
  static char buf[BUF], *bufS, *bufT;
  if (bufS == bufT) {
    bufT = (bufS = buf) + fread(buf, 1, BUF, stdin);
    if (bufS == bufT)
      return EOF;
  }
  return *bufS++;
}
inline int read() {
  int f = 0;
  bool b = 1;
  char ch;
  while (!isdigit(ch = getChar()))
    if (ch == 45)
      b = 0;
  do f = f * 10 + (ch ^ 48);
  while (isdigit(ch = getChar()));
  return b ? f : ~f + 1;
}

const int maxN = 100003, maxE = 1000003, S = maxN - 2, T = maxN - 1;
const long long inf = 0x3f3f3f3f3f3f3f3f;
int cntNode;

int h[maxN], cntE = 1;
struct Edge {
  int tar, val, cost, nxt;
  Edge() {}
  Edge(int tar, int val, int cost, int nxt)
    : tar(tar), val(val), cost(cost), nxt(nxt) {}
}ed[maxE];
inline void addEdge(int x, int y, int z, int w) {
  ed[++cntE] = Edge(y, z, w, h[x]), h[x] = cntE;
  ed[++cntE] = Edge(x, 0, -w, h[y]), h[y] = cntE;
}

int cur[maxN], pre[maxN], maxCost;
inline bool SPFA() {
  static int dis[maxN], que[maxN];
  static bool inq[maxN];
  int hd = 1, tl = 0, i, x, y;
  memset(dis + 1, -0x3f, sizeof(int) * cntNode);
  memcpy(cur + 1, h + 1, sizeof(int) * cntNode);
  dis[S] = 0, dis[T] = -inf, cur[S] = h[S], que[++tl] = S, inq[S] = 1;
  while (hd <= tl) {
    x = que[hd++], inq[x] = 0;
    for (i = h[x]; i; i = ed[i].nxt)
      if (ed[i].val && dis[y = ed[i].tar] < dis[x] + ed[i].cost) {
        dis[y] = dis[x] + ed[i].cost, pre[y] = x;
        if (!inq[y])
          que[++tl] = y, inq[y] = 1;
      }
  }
  return dis[T] > -inf;
}
int dfs(int x, int flow) {
  if (x == T)
    return flow;
  int i, y, k, res = flow;
  for (i = cur[x]; i && res; i = ed[i].nxt) {
    if (ed[i].val && pre[y = ed[i].tar] == x
        && (k = dfs(y, min(res, ed[i].val))))
      ed[i].val -= k, ed[i ^ 1].val += k, res -= k, maxCost += ed[i].cost * k;
    cur[x] = i;
  }
  return flow - res;
}
int maxFlow;
inline void Dinic() {
  while (SPFA())
    maxFlow += dfs(S, inf);
}

signed main() {
  int n = read(), i, x, y, z;
  int p[4] = {2 * n + 1, 2 * n + 2, 2 * n + 3, 2 * n + 4};
  cntNode = 2 * n + 4;
  for (i = 1; i <= n; ++i) {
    x = read(), y = read(), z = read(), addEdge(S, i, z, 0);
    addEdge(i, p[0], inf, x + y), addEdge(i, p[1], inf, -x - y);
    addEdge(i, p[2], inf, x - y), addEdge(i, p[3], inf, y - x);
  }
  for (i = 1; i <= n; ++i) {
    x = read(), y = read(), z = read(), addEdge(i + n, T, z, 0);
    addEdge(p[0], i + n, inf, -x - y), addEdge(p[1], i + n, inf, x + y);
    addEdge(p[2], i + n, inf, y - x), addEdge(p[3], i + n, inf, x - y);
  }
  
  Dinic(), printf("%lld\n", maxCost);
  return 0;
}

### Prompt

In cpp, your task is to solve 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
// Written by newbiechd
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#define int long long
using namespace std;

const int BUF = 1000003;
inline char getChar() {
  static char buf[BUF], *bufS, *bufT;
  if (bufS == bufT) {
    bufT = (bufS = buf) + fread(buf, 1, BUF, stdin);
    if (bufS == bufT)
      return EOF;
  }
  return *bufS++;
}
inline int read() {
  int f = 0;
  bool b = 1;
  char ch;
  while (!isdigit(ch = getChar()))
    if (ch == 45)
      b = 0;
  do f = f * 10 + (ch ^ 48);
  while (isdigit(ch = getChar()));
  return b ? f : ~f + 1;
}

const int maxN = 100003, maxE = 1000003, S = maxN - 2, T = maxN - 1;
const long long inf = 0x3f3f3f3f3f3f3f3f;
int cntNode;

int h[maxN], cntE = 1;
struct Edge {
  int tar, val, cost, nxt;
  Edge() {}
  Edge(int tar, int val, int cost, int nxt)
    : tar(tar), val(val), cost(cost), nxt(nxt) {}
}ed[maxE];
inline void addEdge(int x, int y, int z, int w) {
  ed[++cntE] = Edge(y, z, w, h[x]), h[x] = cntE;
  ed[++cntE] = Edge(x, 0, -w, h[y]), h[y] = cntE;
}

int cur[maxN], pre[maxN], maxCost;
inline bool SPFA() {
  static int dis[maxN], que[maxN];
  static bool inq[maxN];
  int hd = 1, tl = 0, i, x, y;
  memset(dis + 1, -0x3f, sizeof(int) * cntNode);
  memcpy(cur + 1, h + 1, sizeof(int) * cntNode);
  dis[S] = 0, dis[T] = -inf, cur[S] = h[S], que[++tl] = S, inq[S] = 1;
  while (hd <= tl) {
    x = que[hd++], inq[x] = 0;
    for (i = h[x]; i; i = ed[i].nxt)
      if (ed[i].val && dis[y = ed[i].tar] < dis[x] + ed[i].cost) {
        dis[y] = dis[x] + ed[i].cost, pre[y] = x;
        if (!inq[y])
          que[++tl] = y, inq[y] = 1;
      }
  }
  return dis[T] > -inf;
}
int dfs(int x, int flow) {
  if (x == T)
    return flow;
  int i, y, k, res = flow;
  for (i = cur[x]; i && res; i = ed[i].nxt) {
    if (ed[i].val && pre[y = ed[i].tar] == x
        && (k = dfs(y, min(res, ed[i].val))))
      ed[i].val -= k, ed[i ^ 1].val += k, res -= k, maxCost += ed[i].cost * k;
    cur[x] = i;
  }
  return flow - res;
}
int maxFlow;
inline void Dinic() {
  while (SPFA())
    maxFlow += dfs(S, inf);
}

signed main() {
  int n = read(), i, x, y, z;
  int p[4] = {2 * n + 1, 2 * n + 2, 2 * n + 3, 2 * n + 4};
  cntNode = 2 * n + 4;
  for (i = 1; i <= n; ++i) {
    x = read(), y = read(), z = read(), addEdge(S, i, z, 0);
    addEdge(i, p[0], inf, x + y), addEdge(i, p[1], inf, -x - y);
    addEdge(i, p[2], inf, x - y), addEdge(i, p[3], inf, y - x);
  }
  for (i = 1; i <= n; ++i) {
    x = read(), y = read(), z = read(), addEdge(i + n, T, z, 0);
    addEdge(p[0], i + n, inf, -x - y), addEdge(p[1], i + n, inf, x + y);
    addEdge(p[2], i + n, inf, y - x), addEdge(p[3], i + n, inf, x - y);
  }
  
  Dinic(), printf("%lld\n", maxCost);
  return 0;
}

```