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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Threading.Tasks;

public class Program
{
    public static void Main()
    {
        string[] s = Console.ReadLine().Split(' ');
        int N = int.Parse(s[0]);
        int W = int.Parse(s[1]);
        int[,] wall = new int[N, W];

        int[,] c = new int[10, 10];
        for (int i = 0; i < 10; i++)
        {
            s = Console.ReadLine().Split(' ');
            for (int j = 0; j < 10; j++) c[i, j] = int.Parse(s[j]);
        }

        for (int i = 0; i < N; i++)
        {
            s = Console.ReadLine().Split(' ');
            for (int j = 0; j < W; j++)
            {
                wall[i, j] = int.Parse(s[j]);
            }
        }

        int power = 0;
        int p_box = 0;
        int p_pos = 0;
        int rim = 0;

        int[] box = new int[10];
        for (int i = 0; i < 10; i++) box[i] = 0;
        int b = 0;
        int min = 0;

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < W; j++)
            {
                if (wall[i, j] == 1 || wall[i, j] == -1) continue;
                if (box[wall[i, j]] != 0)
                {
                    //Console.WriteLine(wall[i, j] + "を呼び出し");
                    power += box[wall[i, j]];
                    continue;
                }
                else
                {
                    //Console.WriteLine(wall[i, j] + "を記録");
                    b = wall[i, j];
                    p_box = 0;
                    rim = c[wall[i, j], 1];
                    while (true)
                    {
                        min = 10;
                        for (int k = 0; k < 10; k++)
                        {
                            if (min > c[wall[i, j], k] && wall[i, j] != k) { min = c[wall[i, j], k]; p_pos = k; }
                        }
                        p_box += c[wall[i, j], p_pos];
                        wall[i, j] = p_pos;
                        if (rim < p_box) { p_box = rim + 1; break; }
                        if (wall[i, j] == 1) break;
                    }
                    if (rim > p_box) { power += p_box; box[b] = p_box; }
                    else { power += rim; box[b] = rim; }
                }
            }
        }
        Console.WriteLine(power);
    }
}