Problem:
Score : 400 points

Problem Statement

Joisino the magical girl has decided to turn every single digit that exists on this world into 1.

Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points).

She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive).

You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows:

If A_{i,j}≠-1, the square contains a digit A_{i,j}.

If A_{i,j}=-1, the square does not contain a digit.

Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.

Constraints

1≤H,W≤200

1≤c_{i,j}≤10^3 (i≠j)

c_{i,j}=0 (i=j)

-1≤A_{i,j}≤9

All input values are integers.

There is at least one digit on the wall.

Input

Input is given from Standard Input in the following format:

H W
c_{0,0} ... c_{0,9}
:
c_{9,0} ... c_{9,9}
A_{1,1} ... A_{1,W}
:
A_{H,1} ... A_{H,W}

Output

Print the minimum total amount of MP required to turn every digit on the wall into 1 in the end.

Sample Input 1

2 4
0 9 9 9 9 9 9 9 9 9
9 0 9 9 9 9 9 9 9 9
9 9 0 9 9 9 9 9 9 9
9 9 9 0 9 9 9 9 9 9
9 9 9 9 0 9 9 9 9 2
9 9 9 9 9 0 9 9 9 9
9 9 9 9 9 9 0 9 9 9
9 9 9 9 9 9 9 0 9 9
9 9 9 9 2 9 9 9 0 9
9 2 9 9 9 9 9 9 9 0
-1 -1 -1 -1
8 1 1 8

Sample Output 1

12

To turn a single 8 into 1, it is optimal to first turn 8 into 4, then turn 4 into 9, and finally turn 9 into 1, costing 6 MP.

The wall contains two 8s, so the minimum total MP required is 6×2=12.

Sample Input 2

5 5
0 999 999 999 999 999 999 999 999 999
999 0 999 999 999 999 999 999 999 999
999 999 0 999 999 999 999 999 999 999
999 999 999 0 999 999 999 999 999 999
999 999 999 999 0 999 999 999 999 999
999 999 999 999 999 0 999 999 999 999
999 999 999 999 999 999 0 999 999 999
999 999 999 999 999 999 999 0 999 999
999 999 999 999 999 999 999 999 0 999
999 999 999 999 999 999 999 999 999 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1

Sample Output 2

0

Note that she may not need to change any digit.

Sample Input 3

3 5
0 4 3 6 2 7 2 5 3 3
4 0 5 3 7 5 3 7 2 7
5 7 0 7 2 9 3 2 9 1
3 6 2 0 2 4 6 4 2 3
3 5 7 4 0 6 9 7 6 7
9 8 5 2 2 0 4 7 6 5
5 4 6 3 2 3 0 5 4 3
3 6 2 3 4 2 4 0 8 9
4 6 5 4 3 5 3 2 0 8
2 1 3 4 5 7 8 6 4 0
3 5 2 6 1
2 5 3 2 1
6 9 2 5 6

Sample Output 3

47

Solution in Python (3.4.3) (judge verdict: Accepted)

def solve(inputstr):
  return inputstr

def rec_cost(x,acc,plist,clist):
  if x == 1:
    return acc
  else:
    minimum = -1
    for next in plist:
      nextplist = plist.copy()
      nextplist.remove(next)
      candidate = rec_cost(next,acc+clist[x][next],nextplist,clist)
      if minimum == -1:
        minimum = candidate
      elif candidate < minimum:
        minimum = candidate
    return minimum
      

def cost(n,clist):
 # print("invoked")
  plist = [0,1,2,3,4,5,6,7,8,9]
  plist.remove(n)
  return rec_cost(n,0,plist,clist)

def joisino(a,clist):
  note = [-1,0,-1,-1,-1,-1,-1,-1,-1,-1]
  answer = 0
  for raw in a:
    for number in raw:
      if number != -1:
        if note[number] == -1:
          note[number] = cost(number,clist)
        answer += note[number]
      
  return answer

def i_o():
  h, w = map(int,input().split(" "))
  i = 0 
  clist = []
  while i < 10:
    raw = list(map(int,input().split(" ")))
    clist.append(raw)
    i += 1
  a = []
  i = 1
  while i <= h:
    raw = list(map(int,input().split(" ")))
    a.append(raw)
    i += 1
  return h,w,clist,a

def fact(n):
  if n == 1:
    return 1
  else:
    return n * fact(n-1)
  
  
 
if __name__ == '__main__':
  h,w,clist,a = i_o()
  ans =joisino(a,clist)
  
  print(ans, flush=True)