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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main{

  public static void main(String[] args){
      Scanner sc = new Scanner(System.in);
      int h = sc.nextInt();
      int w  = sc.nextInt();

      int[][] costs = new int[10][10];
      for(int i = 0; i <= 9; i++) {
        for(int j = 0; j <= 9; j++)
          costs[i][j] = sc.nextInt();
      }

      int[][] matrix = new int[h][w];
      for(int i = 0; i < h; i++) {
          for(int j = 0; j < w; j++) {
            matrix[i][j] = sc.nextInt();
          }
      }
      
      Map<Integer, Long> memo = new HashMap<>();
      long cost = 0;
      for(int i = 0; i < h; i++) {
          for(int j = 0; j < w; j++) {
              int num = matrix[i][j]; // 8
              if(num == -1) continue; //       8            9
              long minCost = 0;
              if(memo.containsKey(num))
                minCost = memo.get(num);
              else{
                minCost = dfs(memo, costs, matrix, num, 0, costs[num][1]);
                memo.put(num, minCost);
              }
              cost += minCost;
          }
      }
      System.out.println(cost);
  }

  private static long dfs(Map<Integer, Long> memo, int[][] costs, int[][] matrix, int n, long curCost, long minCost) {
      
      if(curCost >= minCost) return minCost;
      if(n == 1) {
        return curCost;
      }

      for(int i = 0; i <= 9; i++) {
        if(i == n) continue;
        minCost = Math.min(minCost, dfs(memo, costs, matrix, i, curCost + costs[n][i], minCost));
      }
      //memo.put(n, minCost);
      return minCost;
  }
}