#include "iostream"
#include "algorithm"
using namespace std;
structEdge
{
    int begin;
    int end;
    int length;
};
Edge Num[10001];
int Father[1001], Map[1001][1001]; 
int Minlen, Edge_Num,i, j,TmpBegin, TmpEnd, counter, Total, BuildNum, From, To;
int Find(int k)
{
    return Father[k]==k?k:Father[k]=Find(Father[k]);
}
int Cmp(Edge a, Edge b)
{
    return a.length<b.length;
}
int Kruskal()
{
    Minlen = 0;
    Edge_Num = (Total-1)*Total/2;
    for(i=0; i<Edge_Num; i++)
    {
        TmpBegin = Find(Num[i].begin);
        TmpEnd = Find(Num[i].end);
        if(TmpBegin != TmpEnd)
        {
            Father[TmpBegin] = TmpEnd;
            Minlen += Num[i].length;
        }
    }
    return Minlen;
}
void Init()
{
    cin>>Total;
    counter = 0;
    Edge_Num = (Total-1)*Total/2;
    for(i=1; i<=Total; i++)
    {
        for(j=1; j<=Total; j++)
        {
            cin>>Map[i][j];
            if(j>i)
            {
                Num[counter].begin = i;
                Num[counter].end = j;
                Num[counter++].length = Map[i][j];
            }
        }
        Father[i] = i;
    }
    cin>>BuildNum;
    while(BuildNum--)
    {
        cin>>From>>To;
        TmpBegin = Find(From);  //1
        TmpEnd = Find(To);  //2
        Father[TmpBegin] = TmpEnd;  //3 這個地方就是將已經存在的邊用并查集給合并起來，這就是本題的關鍵了。
              //標有1 2 3的這三行就是用來處理已經建好的道路的
    }
    sort(Num, Num+counter, Cmp);
}
int main()
{
    Init();
    cout<<Kruskal()<<endl;
    return 0;
}