#include <bits/stdc++.h>
//#define fastIO cin.tie(0);ios_base::sync_with_stdio(false)
#define N 10010
#define ll long long
#define INF 1000000000000000000
//min-cost flow problem
using namespace std;

struct edge {
  ll from, to, cap, flow, cost;
};

ll n, d[2 * N + 5], pre[2 * N + 5], Maxflow = 0, COST = 0;
bool inqueue[2 * N + 5];
vector<edge> Edge;     // store edges
vector<ll> G[2 * N + 10];

inline void addEdge( ll from, ll to, ll cap, ll cost )
{
    G[from].push_back( Edge.size() );
    Edge.push_back( edge{from, to, cap, 0, cost} );
    
    G[to].push_back( Edge.size() );
    Edge.push_back( edge{to, from, 0, 0, -cost} );
}

inline bool bfs( ll s, ll t )
{
    for ( int i = 0; i < 2 * N + 5; ++i )
        d[i] = INF;
    memset( inqueue, 0, sizeof( inqueue ) );

    queue<ll> q;
    q.push( s );

    d[s] = 0, inqueue[s] = true;

    ll now;
    while ( q.empty() == false ) {
        now = q.front();
        q.pop();
        inqueue[now] = false;
        for ( ll i = 0; i < G[now].size(); ++i ) {
            edge& tmp = Edge[G[now][i]];
            if ( tmp.cap > tmp.flow && d[now] + tmp.cost < d[tmp.to] ) {
                d[tmp.to] = d[now] + tmp.cost;
                pre[tmp.to] = G[now][i];

                if ( inqueue[tmp.to] == false && tmp.to != t ) {
                    inqueue[tmp.to] = true;
                    q.push( tmp.to );
                }
            }
        }
    }

    return d[t] != INF;
    /*
    if(d[t]==-INF)
        return false;
    else
        return true;
    */
}

inline void calc( ll s, ll t )
{
    ll minflow = INF;
    ll tmpu = t;
    while ( s != tmpu ) { //find minflow
        minflow = min( minflow, Edge[pre[tmpu]].cap - Edge[pre[tmpu]].flow );
        tmpu = Edge[pre[tmpu]].from;
    }

    Maxflow += minflow;
    COST += minflow * d[t];

    while ( s != t ) {
        Edge[pre[t]].flow += minflow;
        Edge[pre[t] ^ 1].flow -= minflow;
        t = Edge[pre[t]].from;
    }
}

inline void init()
{
    cin >> n;

    ll a, b, c;
    for ( int i = 1; i <= n; ++i ) {
        cin >> a >> b >> c;
        addEdge( 0, i, c, 0 );
        addEdge( i, 2 * n + 1, c, -a - b );
        addEdge( i, 2 * n + 2, c, a - b );
        addEdge( i, 2 * n + 3, c, -a + b );
        addEdge( i, 2 * n + 4, c, a + b );
    }
    for ( int i = 1; i <= n; ++i ) {
        cin >> a >> b >> c;
        addEdge( i + n, 2 * n + 5, c, 0 );
        addEdge( 2 * n + 1, i + n, c, a + b );
        addEdge( 2 * n + 2, i + n, c, -a + b );
        addEdge( 2 * n + 3, i + n, c, a - b );
        addEdge( 2 * n + 4, i + n, c, -a - b );
    }
}

inline void solve()
{
    ll s = 0, t = 2 * n + 5;

    while ( bfs( s, t ) == true ) {
        calc( s, t );  //update cost && Maxflow
    }
    cout << -COST << endl;
}

int main()
{
    init();
    solve();
}
### Prompt

Develop a solution in cpp 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>
//#define fastIO cin.tie(0);ios_base::sync_with_stdio(false)
#define N 10010
#define ll long long
#define INF 1000000000000000000
//min-cost flow problem
using namespace std;

struct edge {
  ll from, to, cap, flow, cost;
};

ll n, d[2 * N + 5], pre[2 * N + 5], Maxflow = 0, COST = 0;
bool inqueue[2 * N + 5];
vector<edge> Edge;     // store edges
vector<ll> G[2 * N + 10];

inline void addEdge( ll from, ll to, ll cap, ll cost )
{
    G[from].push_back( Edge.size() );
    Edge.push_back( edge{from, to, cap, 0, cost} );
    
    G[to].push_back( Edge.size() );
    Edge.push_back( edge{to, from, 0, 0, -cost} );
}

inline bool bfs( ll s, ll t )
{
    for ( int i = 0; i < 2 * N + 5; ++i )
        d[i] = INF;
    memset( inqueue, 0, sizeof( inqueue ) );

    queue<ll> q;
    q.push( s );

    d[s] = 0, inqueue[s] = true;

    ll now;
    while ( q.empty() == false ) {
        now = q.front();
        q.pop();
        inqueue[now] = false;
        for ( ll i = 0; i < G[now].size(); ++i ) {
            edge& tmp = Edge[G[now][i]];
            if ( tmp.cap > tmp.flow && d[now] + tmp.cost < d[tmp.to] ) {
                d[tmp.to] = d[now] + tmp.cost;
                pre[tmp.to] = G[now][i];

                if ( inqueue[tmp.to] == false && tmp.to != t ) {
                    inqueue[tmp.to] = true;
                    q.push( tmp.to );
                }
            }
        }
    }

    return d[t] != INF;
    /*
    if(d[t]==-INF)
        return false;
    else
        return true;
    */
}

inline void calc( ll s, ll t )
{
    ll minflow = INF;
    ll tmpu = t;
    while ( s != tmpu ) { //find minflow
        minflow = min( minflow, Edge[pre[tmpu]].cap - Edge[pre[tmpu]].flow );
        tmpu = Edge[pre[tmpu]].from;
    }

    Maxflow += minflow;
    COST += minflow * d[t];

    while ( s != t ) {
        Edge[pre[t]].flow += minflow;
        Edge[pre[t] ^ 1].flow -= minflow;
        t = Edge[pre[t]].from;
    }
}

inline void init()
{
    cin >> n;

    ll a, b, c;
    for ( int i = 1; i <= n; ++i ) {
        cin >> a >> b >> c;
        addEdge( 0, i, c, 0 );
        addEdge( i, 2 * n + 1, c, -a - b );
        addEdge( i, 2 * n + 2, c, a - b );
        addEdge( i, 2 * n + 3, c, -a + b );
        addEdge( i, 2 * n + 4, c, a + b );
    }
    for ( int i = 1; i <= n; ++i ) {
        cin >> a >> b >> c;
        addEdge( i + n, 2 * n + 5, c, 0 );
        addEdge( 2 * n + 1, i + n, c, a + b );
        addEdge( 2 * n + 2, i + n, c, -a + b );
        addEdge( 2 * n + 3, i + n, c, a - b );
        addEdge( 2 * n + 4, i + n, c, -a - b );
    }
}

inline void solve()
{
    ll s = 0, t = 2 * n + 5;

    while ( bfs( s, t ) == true ) {
        calc( s, t );  //update cost && Maxflow
    }
    cout << -COST << endl;
}

int main()
{
    init();
    solve();
}
```