#include <bits/stdc++.h>

using namespace std;
using ll=long long;
using pi=pair<int, ll>;
const int inf = 0x3f3f3f3f;

const int V = 2048;
const int E = (V*5)<<2;
int sson[V], pre[V], trk[V]; ll dis[V];
int cnt;

struct node{
    int to, bro, cap, flow; ll cost;
} edge[E];

void init(){
    cnt = 0;
    memset(sson, -1, sizeof sson);
}

void add_edge(int u, int v, int cap, ll cost){
    edge[cnt] = {v, sson[u], cap, 0, cost};
    sson[u] = cnt++;
    edge[cnt] = {u, sson[v], 0, 0, -cost};
    sson[v] = cnt++;
}

bool spfa(int s, int t){
    queue<int> q;
    memset(pre, -1, sizeof pre);
    memset(dis, inf, sizeof dis);
    memset(trk, 0, sizeof trk);
    dis[s] = 0; trk[s] = 1; q.push(s);
    while(!q.empty()){
        int u = q.front(); q.pop(); trk[u] = 0;
        for (int e=sson[u]; ~e; e=edge[e].bro){
            int v = edge[e].to;
            if (edge[e].cap > edge[e].flow && dis[v] > dis[u] + edge[e].cost){
                dis[v] = dis[u] + edge[e].cost;
                pre[v] = e;
                if (!trk[v]){
                    trk[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}

pi mcmf(int s, int t){
    int flow = 0; ll cost = 0;
    while (spfa(s, t)){
        int f = inf;
        for (int e=pre[t]; ~e; e=pre[edge[e^1].to])
            f = min(f, edge[e].cap - edge[e].flow);
        for (int e=pre[t]; ~e; e=pre[edge[e^1].to]){
            edge[e].flow += f;
            edge[e^1].flow -= f;
            cost += edge[e].cost * f;
        }
        flow += f;
    }
    return {flow, cost};
}
int dx[4] = {1, -1, 1, -1};
int dy[4] = {1, -1, -1, 1};
void solve() {
    int n; cin >> n;
    init();
    int S = 4, T = 5;
    int v = 6;
    for (int _ = 0; _ < n; _++) {
        int x,y,z;
        cin >> x >> y >> z;
        add_edge(S, v, z, 0);
        for (int i = 0; i < 4; i++) {
            ll cost = x*dx[i] + y*dy[i];
            add_edge(v, i, z, cost);
        }
        v++;
    }
    for (int _ = 0; _ < n; _++) {
        int x,y,z;
        cin >> x >> y >> z;
        add_edge(v, T, z, 0);
        for (int i = 0; i < 4; i++) {
            ll cost = x*dx[i^1] + y*dy[i^1];
            add_edge(i, v, z, cost);
        }
        v++;
    }

    ll res = -mcmf(S,T).second;
    cout << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

### Prompt

Your task is to create 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 <bits/stdc++.h>

using namespace std;
using ll=long long;
using pi=pair<int, ll>;
const int inf = 0x3f3f3f3f;

const int V = 2048;
const int E = (V*5)<<2;
int sson[V], pre[V], trk[V]; ll dis[V];
int cnt;

struct node{
    int to, bro, cap, flow; ll cost;
} edge[E];

void init(){
    cnt = 0;
    memset(sson, -1, sizeof sson);
}

void add_edge(int u, int v, int cap, ll cost){
    edge[cnt] = {v, sson[u], cap, 0, cost};
    sson[u] = cnt++;
    edge[cnt] = {u, sson[v], 0, 0, -cost};
    sson[v] = cnt++;
}

bool spfa(int s, int t){
    queue<int> q;
    memset(pre, -1, sizeof pre);
    memset(dis, inf, sizeof dis);
    memset(trk, 0, sizeof trk);
    dis[s] = 0; trk[s] = 1; q.push(s);
    while(!q.empty()){
        int u = q.front(); q.pop(); trk[u] = 0;
        for (int e=sson[u]; ~e; e=edge[e].bro){
            int v = edge[e].to;
            if (edge[e].cap > edge[e].flow && dis[v] > dis[u] + edge[e].cost){
                dis[v] = dis[u] + edge[e].cost;
                pre[v] = e;
                if (!trk[v]){
                    trk[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}

pi mcmf(int s, int t){
    int flow = 0; ll cost = 0;
    while (spfa(s, t)){
        int f = inf;
        for (int e=pre[t]; ~e; e=pre[edge[e^1].to])
            f = min(f, edge[e].cap - edge[e].flow);
        for (int e=pre[t]; ~e; e=pre[edge[e^1].to]){
            edge[e].flow += f;
            edge[e^1].flow -= f;
            cost += edge[e].cost * f;
        }
        flow += f;
    }
    return {flow, cost};
}
int dx[4] = {1, -1, 1, -1};
int dy[4] = {1, -1, -1, 1};
void solve() {
    int n; cin >> n;
    init();
    int S = 4, T = 5;
    int v = 6;
    for (int _ = 0; _ < n; _++) {
        int x,y,z;
        cin >> x >> y >> z;
        add_edge(S, v, z, 0);
        for (int i = 0; i < 4; i++) {
            ll cost = x*dx[i] + y*dy[i];
            add_edge(v, i, z, cost);
        }
        v++;
    }
    for (int _ = 0; _ < n; _++) {
        int x,y,z;
        cin >> x >> y >> z;
        add_edge(v, T, z, 0);
        for (int i = 0; i < 4; i++) {
            ll cost = x*dx[i^1] + y*dy[i^1];
            add_edge(i, v, z, cost);
        }
        v++;
    }

    ll res = -mcmf(S,T).second;
    cout << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

```