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: Accepted)

#
# 優先度つきキュー + ダイクストラ
# ダイクストラなので負の辺は不可
#
class PriorityQueue
  def initialize
    @q = []
    @h = {}
  end

  # push obj unless already contained
  def push_ex(obj, cost)
    return if @h.key?(obj)
    i = (0...@q.size).bsearch{|i| @q[i][1] > cost} || @q.size
    @q.insert(i, [obj, cost])
    @h[obj] = true
  end

  def pop
    return nil if @q.empty?
    obj = @q.shift[0]
    @h.delete(obj)
    return obj
  end

  # routeはデバッグ用
  def self.dijkstra_with_cost(start)
    best = Array.new(10){ [Float::INFINITY, nil] }
    best[start] = [0, [start]]
    q = PriorityQueue.new
    q.push_ex([start, [start]], 0)
    while (item = q.pop)
      a, route = *item
      10.times do |b|
        next if b == a
        cost = COST[b][a]
        new_cost = best[a][0] + cost
        #p a: a, b: b, route: route, cost: cost, best: best[b][0], new_cost: new_cost
        if new_cost < best[b][0]
          best[b] = [new_cost, route + [b]]
          q.push_ex([b, route+[b]], new_cost)
        end
      end
      #p best: best
    end
    return best
  end
end

H, W = gets.split.map(&:to_i)
COST = 10.times.map{ gets.split.map(&:to_i) }
NUMS = $stdin.read.split.map(&:to_i)

cost = PriorityQueue.dijkstra_with_cost(1)
cost.push([0, nil])  # for -1
#pp cost: cost

p NUMS.map{|n| cost[n][0]}.inject(:+)