#include<iostream>

 #include<vector>

 #include<cstring>

 #include<cstdio>

 #include<queue>

 using namespace std;

 const int INF = 0x3f3f3f3f;

 const int maxn =  + ;//顾客的数目

 const int maxm =  + ;//猪圈的数目

 struct edge

 {

     int u, v, c, f;

     edge(int u, int v, int c, int f):u(u), v(v), c(c), f(f){}

 };

 vector<edge>e;

 vector<int>G[maxn];

 int a[maxn], p[maxn];

 int m;

 void init(int n)

 {

     for(int i = ; i <= n; i++)

         G[i].clear();

     e.clear();

 }

 void addedge(int u, int v, int c)

 {

     e.push_back(edge(u, v, c, ));

     e.push_back(edge(v, u, , ));//反向边

     m = e.size();

     G[u].push_back(m - );

     G[v].push_back(m - );

 }

 int Maxflow(int s, int t)

 {

     int flow = ;

     for(;;)

     {

         memset(a, , sizeof(a));//每次找增广路的每个点的流量

         queue<int>q;

         q.push(s);

         a[s] = INF;

         while(!q.empty())

         {

             int u = q.front();

             q.pop();

             for(int i = ; i < G[u].size(); i++)

             {

                 edge& now = e[G[u][i]];

                 int v = now.v;

                 if(!a[v] && now.c > now.f)//v点还没有流到并且这条边还没满

                 {

                     p[v] = G[u][i];//反向记录边

                     a[v] = min(a[u], now.c - now.f);//到达v点的水量取决于边剩余的容量和u点的水量

                     q.push(v);

                 }

             }

             if(a[t])break;//已到达终点

         }

         if(!a[t])break;//已经找不到增广路

         for(int u = t; u != s; u = e[p[u]].u)

         {

             e[p[u]].f += a[t];

             e[p[u]^].f -= a[t];

         }

         flow += a[t];

     }

     return flow;

 }

 int pig[maxm];//每个猪圈猪的数目

 int last[maxm];//每个猪圈的前一个顾客的序号

 int Map[maxn][maxn];//顾客顾客之间的边，有重边，先用数组记录

 int main()

 {

     int n, m;

     cin >> m >> n;

     for(int i = ; i <= m; i++)cin >> pig[i];

     int s = , t = n + ;//设置源点和汇点

     int d, x;

     for(int i = ; i <= n; i++)//顾客编号

     {

         cin >> d;

         for(int j = ; j < d; j++)

         {

             cin >> x;//猪圈编号

             if(last[x] == )//猪圈的第一个顾客

                 addedge(s, i, pig[x]);

             else addedge(last[x], i, INF);//顾客i紧接着顾客last[x]后面打开猪圈x

             last[x] = i;

         }

         cin >> x;//到汇点的边的权值

         addedge(i, t, x);

     }

     cout<<Maxflow(s, t)<<endl;

     return ;

 }