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 Ruby (2.3.3) (judge verdict: Runtime Error)

def getmin(arr,alist)
  rsco = 2000
  ridx = alist[0]
  check = false
  alist.each do |idx|
    if(rsco > arr[idx] && arr[idx] != 0)
      rsco = arr[idx]
      ridx = idx
      check = true
    end
  end
  return ridx,rsco if check
  return ridx,0
end
hw = Hash.new(0)
cost = []
width = 3
result = 0
onecost = []
s = gets.chomp.split(/\s+/)
hw[:h] = s[0].to_i
hw[:w] = s[1].to_i
width.times do |i|
  cost[i] = []
  s = gets.chomp.split(/\s+/)
  width.times do |ii|
    cost[i][ii] = s[ii].to_i
  end
end
#探索
width.times do |i|
  d = []
  width.times do |ij|
    d[ij] = cost[i][ij]
  end
  alist = [i]
  olist = []
  unfinish = true
  while(unfinish)
    v,s = getmin(d,alist)
    if (v == 1)
      onecost[i] = s
      unfinish = false
    else
      width.times do |w|
        next if (w == v || olist.include?(w))
        alist.push(w) unless alist.include?(w)
        if(d[w] > (d[v] + cost[v][w]))
          d[w] = d[v] + cost[v][w]
        end
      end
    end
    alist.delete(v)
    olist.push(v)
  end
end
hw[:h].times do |h|
  s = gets.chomp.split(/\s+/)
  hw[:w].times do |w|
    num = s[w].to_i
    next if (num == -1 || num ==1)
    result += onecost[num]
  end
end
print result,"\n"