//最小生成树裸题，稠密图，用prim实现不用kruskal

#include 
     
      #include 
      
       #define MAX 110#define INF 1000*100+10int g[MAX][MAX];int n,m;int lowcoat[MAX],adj[MAX];void Prim(){    int i,j,k,min,ans;    for(i=1; i<=n; i++)        lowcoat[i]=g[1][i] , adj[i]=1;    lowcoat[1]=0;    for(i=1; i
       
         g[k][j])                lowcoat[j]=g[k][j] , adj[j]=k;    }    for(ans=0 , i=1; i<=n; i++)        if(g[i][j=adj[i]]!=-1)             ans+=g[i][j];     printf("%d\n",ans);    return ;}int main(){    int i,j,a,b;    while(scanf("%d",&n)!=EOF)    {        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                scanf("%d",&g[i][j]);        scanf("%d",&m);        for(i=1; i<=m; i++)        {            scanf("%d%d",&a,&b);            g[a][b]=g[b][a]=-1;  //不能改为0，会导致算法出错，仔细想想即可        }/*        for(i=1; i<=n; i++)        {            for(j=1; j<=n; j++)                printf("%d ",g[i][j]);            printf("\n");        }*/        Prim();    }    return 0;}