/*

     题意：上司在，员工不在，反之不一定。每一个人有一个权值，问权值和最大多少。

     树形DP：把上司和员工的关系看成根节点和子节点的关系，两者有状态转移方程：

             dp[rt][0] += max (dp[son][1], dp[son][0]);    //上司不去

             dp[rt][1] += dp[son][0];                    //上司去，员工都不去

 */

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 #include <cmath>

 using namespace std;

 const int MAXN = 6e3 + ;

 const int INF = 0x3f3f3f3f;

 bool vis[MAXN];

 vector<int> edge[MAXN];

 int dp[MAXN][];

 int n;

 void DFS(int rt)

 {

     vis[rt] = true;

     dp[rt][] = ;

     for (int i=; i<edge[rt].size (); ++i)

     {

         int son = edge[rt][i];

         if (!vis[son])

         {

             DFS (son);

             dp[rt][] += max (dp[son][], dp[son][]);

             dp[rt][] += dp[son][];

         }

     }

 }

 int main(void)        //URAL 1039 Anniversary Party

 {

 //    freopen ("URAL_1039.in", "r", stdin);

     while (scanf ("%d", &n) == )

     {

         memset (vis, false, sizeof (vis));

         memset (dp, , sizeof (dp));

         for (int i=; i<=n; ++i)    scanf ("%d", &dp[i][]);

         int l, k;

         while (scanf ("%d%d", &l, &k) == )

         {

             if (l ==  && k == )    break;

             vis[l] = true;

             edge[l].push_back (k);

             edge[k].push_back (l);

         }

         int root = ;

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

         {

             if (!vis[i])    {root = i;    break;}

         }

         memset (vis, false, sizeof (vis));    DFS (root);

         printf ("%d\n", max (dp[root][], dp[root][]));

     }

     return ;

 }