#include <bits/stdc++.h>
using namespace std;

const int N = 3010;

typedef long long LL;

struct edge {
    int to, next, w, c;
} e[N * 10];

int head[N], ecnt = 1;

inline void adde(int from, int to, int w, int c) {
    e[++ecnt] = (edge){to, head[from], w, -c}, head[from] = ecnt;
    e[++ecnt] = (edge){from, head[to], 0, c}, head[to] = ecnt;
}

int vis[N], cur[N]; LL dis[N];

inline bool SPFA(int s, int t) {
    memset(dis, 0x3f, sizeof(dis)), memset(vis, 0, sizeof(vis));
    queue<int> q; q.push(t), vis[t] = 1, dis[t] = 0;
    while (!q.empty()) {
        int u = q.front(); q.pop(), vis[u] = 0;
        for (int i = head[u]; i; i = e[i].next)
            if (e[i ^ 1].w && dis[e[i].to] > dis[u] - e[i].c) {
                dis[e[i].to] = dis[u] - e[i].c;
                if (!vis[e[i].to]) q.push(e[i].to), vis[e[i].to] = 1;
            }
    }
    return dis[s] < 1e17;
}

LL ans;

int DFS(int u, int t, int f) {
    if (u == t || !f) return f;
    int res = 0; vis[u] = 1;
    for (int &i = cur[u], tmp; i; i = e[i].next)
        if (!vis[e[i].to] && e[i].w && dis[u] - e[i].c == dis[e[i].to] && 
            (tmp = DFS(e[i].to, t, min(f, e[i].w)))) {
                res += tmp, f -= tmp, e[i].w -= tmp, e[i ^ 1].w += tmp, ans += (LL)tmp * e[i].c;
                if (!f) break;
            }
    return res;
}

inline LL dinic(int s, int t) {
    while (SPFA(s, t)) memcpy(cur, head, sizeof(head)), DFS(s, t, 1e9);
    return ans;
}

int main() {
    int n, s, t; scanf("%d", &n), s = n * 2 + 5, t = s + 1;
    for (int i = 1; i <= n; i++) {
        int x, y, c; scanf("%d%d%d", &x, &y, &c);
        adde(s, i, c, 0);
        adde(i, 2 * n + 1, 1e9, -x - y), adde(i, 2 * n + 2, 1e9, x + y);
        adde(i, 2 * n + 3, 1e9, y - x), adde(i, 2 * n + 4, 1e9, x - y);
    }
    for (int i = 1; i <= n; i++) {
        int x, y, c; scanf("%d%d%d", &x, &y, &c);
        adde(i + n, t, c, 0);
        adde(2 * n + 1, i + n, 1e9, x + y), adde(2 * n + 2, i + n, 1e9, -x - y);
        adde(2 * n + 3, i + n, 1e9, x - y), adde(2 * n + 4, i + n, 1e9, y - x);
    }
    printf("%lld\n", -dinic(s, t));
    return 0;
}//
### Prompt

Please create a solution in Cpp 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 <bits/stdc++.h>
using namespace std;

const int N = 3010;

typedef long long LL;

struct edge {
    int to, next, w, c;
} e[N * 10];

int head[N], ecnt = 1;

inline void adde(int from, int to, int w, int c) {
    e[++ecnt] = (edge){to, head[from], w, -c}, head[from] = ecnt;
    e[++ecnt] = (edge){from, head[to], 0, c}, head[to] = ecnt;
}

int vis[N], cur[N]; LL dis[N];

inline bool SPFA(int s, int t) {
    memset(dis, 0x3f, sizeof(dis)), memset(vis, 0, sizeof(vis));
    queue<int> q; q.push(t), vis[t] = 1, dis[t] = 0;
    while (!q.empty()) {
        int u = q.front(); q.pop(), vis[u] = 0;
        for (int i = head[u]; i; i = e[i].next)
            if (e[i ^ 1].w && dis[e[i].to] > dis[u] - e[i].c) {
                dis[e[i].to] = dis[u] - e[i].c;
                if (!vis[e[i].to]) q.push(e[i].to), vis[e[i].to] = 1;
            }
    }
    return dis[s] < 1e17;
}

LL ans;

int DFS(int u, int t, int f) {
    if (u == t || !f) return f;
    int res = 0; vis[u] = 1;
    for (int &i = cur[u], tmp; i; i = e[i].next)
        if (!vis[e[i].to] && e[i].w && dis[u] - e[i].c == dis[e[i].to] && 
            (tmp = DFS(e[i].to, t, min(f, e[i].w)))) {
                res += tmp, f -= tmp, e[i].w -= tmp, e[i ^ 1].w += tmp, ans += (LL)tmp * e[i].c;
                if (!f) break;
            }
    return res;
}

inline LL dinic(int s, int t) {
    while (SPFA(s, t)) memcpy(cur, head, sizeof(head)), DFS(s, t, 1e9);
    return ans;
}

int main() {
    int n, s, t; scanf("%d", &n), s = n * 2 + 5, t = s + 1;
    for (int i = 1; i <= n; i++) {
        int x, y, c; scanf("%d%d%d", &x, &y, &c);
        adde(s, i, c, 0);
        adde(i, 2 * n + 1, 1e9, -x - y), adde(i, 2 * n + 2, 1e9, x + y);
        adde(i, 2 * n + 3, 1e9, y - x), adde(i, 2 * n + 4, 1e9, x - y);
    }
    for (int i = 1; i <= n; i++) {
        int x, y, c; scanf("%d%d%d", &x, &y, &c);
        adde(i + n, t, c, 0);
        adde(2 * n + 1, i + n, 1e9, x + y), adde(2 * n + 2, i + n, 1e9, -x - y);
        adde(2 * n + 3, i + n, 1e9, x - y), adde(2 * n + 4, i + n, 1e9, y - x);
    }
    printf("%lld\n", -dinic(s, t));
    return 0;
}//
```