These are N cities in Spring country. Between each pair of cities there may be 
one transportation track or none. Now there is some cargo that should be delivered 
from one city to another. The transportation fee consists of two parts:

The cost of the transportation on the path between these cities, and

a certain tax which will be charged whenever any cargo passing through one city, except for the source and the destination cities.

You must write a program to find the route which has the minimum cost.


Input

First is N, number of cities. N = 0 indicates the end of input.

The data of path cost, city tax, source and destination cities are given in the input, which is of the form:

a11 a12 ... a1N
a21 a22 ... a2N
...............
aN1 aN2 ... aNN
b1  b2  ... bN

c d
e f
...
g h

where aij is the transport cost from city i to city j, aij = -1 indicates there is no direct path between city i and city j. bi represents the tax of passing through city i. And the cargo is to be delivered from city c to city d, city e to city f, ..., and g = h = -1. You must output the sequence of cities passed by and the total cost which is of the form:


Output

From c to d :
Path: c-->c1-->......-->ck-->d
Total cost : ......
......

From e to f :
Path: e-->e1-->..........-->ek-->f
Total cost : ......

Note: if there are more minimal paths, output the lexically smallest one. Print a blank line after each test case.


Sample Input

5
0 3 22 -1 4
3 0 5 -1 -1
22 5 0 9 20
-1 -1 9 0 4
4 -1 20 4 0
5 17 8 3 1
1 3
3 5
2 4
-1 -1
0


Sample Output

From 1 to 3 :
Path: 1-->5-->4-->3
Total cost : 21

From 3 to 5 :
Path: 3-->4-->5
Total cost : 16

From 2 to 4 :
Path: 2-->1-->5-->4
Total cost : 17


#define INF 99999999
int Edge[100][100];
int cost[100];
int path[100][100];
int main(){
  int i,j,k,n,m,u,v,len,s,g,w,str,end;
  while(cin>>n&&n){
    for(i=1;i<=n;i++)
     for(j=1;j<=n;j++){
        scanf("%d",&w);
        if(w==-1)
          Edge[i][j]=INF;
        else
          Edge[i][j]=w;
       path[i][j]=j;    //i到j这条路 i的后继结点为j 即使不能走，因为后面会更新
     }
   for(i=1;i<=n;i++)
     scanf("%d",&cost[i]);
   for(k=1;k<=n;k++)
    for(i=1;i<=n;i++)
     for(j=1;j<=n;j++){
        s=Edge[i][k]+Edge[k][j]+cost[k];
        if(s<Edge[i][j]){
          Edge[i][j]=s;
          path[i][j]=path[i][k]; //i到j这条路 i经过k 于是i先到k 所以后继结点变为i到k的后继结点
        }
        else if(s==Edge[i][j]){  //字典序，即可以走k 也可以走i
           if(path[i][k]<path[i][j]) //走k经过的点比走j小
             path[i][j]=path[i][k];
        }
     }
    while(scanf("%d%d",&str,&end)!=EOF&&str!=-1&&end!=-1){
        printf("From %d to %d :\n",str,end);
        printf("Path: %d",str);
        k=str; //起点
       while(k!=end){  //终点是end 依次输出str的后继结点
         printf("-->%d",path[k][end]);
         k=path[k][end];
       }
       printf("\nTotal cost : %d\n\n",Edge[str][end]);
    }
  }
  return 0;
}