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 (2.7.1) (judge verdict: Accepted)

def dijkstra(ini, goal, g)
  node = Array.new(g.length){Array.new(2,-1)} #最小コスト、確定ノードか否か
  node[ini][0] = 0
  node[ini][1] = -1
  while true
    min = -1
    min_i = -1
    g.length.times do |i|
      if node[i][1] == -1
        min_i = i if (min > node[i][0] || min == -1) && node[i][0] != -1
        min = node[i][0] if (min > node[i][0] || min == -1) && node[i][0] != -1
      end
    end
    if min_i == -1
      break
    end
    node[min_i][1] = 1
    g[min_i].each do |x|
      if node[x[0]][1] == -1
        if node[x[0]][0] == -1 || node[x[0]][0] > node[min_i][0] + x[1]
          node[x[0]][0] = node[min_i][0] + x[1]
        end
      end
    end
  end
  node[goal][0]
end

h,w = gets.chomp.split(" ").map(&:to_i)
c = Array.new(10){Array.new(10)}
10.times do |i|
  c[i] = gets.chomp.split(" ").map(&:to_i)
end
a = Array.new(h){Array.new(w)}
h.times do |i|
  a[i] = gets.chomp.split(" ").map(&:to_i)
end
g = Array.new(10){Array.new}
10.times do |i|
  10.times do |j|
    g[i].push([j,c[i][j]]) if i != j
  end
end
mp = []
10.times do |i|
  mp[i] = dijkstra(i,1,g)
end
ans = 0
h.times do |i|
  w.times do |j|
    ans += mp[a[i][j]] if a[i][j] != -1
  end
end
puts ans