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: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;


public class D
{
	public static void Main()
	{
		string[] args;
		
		args = Console.ReadLine().Split(' ');
		int H = Int32.Parse(args[0]);
		int W = Int32.Parse(args[1]);

		int[,] c = new int[10,10];
		for(int i = 0; i < 10; i++)
		{
			args = Console.ReadLine().Split(' ');
			for(int j = 0; j < 10; j++)
			{
				c[i,j] = Int32.Parse(args[j]);
			}
		}

		int[] num = new int[10];
		for(int i = 0; i < H; i++)
		{
			args = Console.ReadLine().Split(' ');
			for(int j = 0; j < W; j++)
			{
				int A = Int32.Parse(args[j]);
				if(A != -1)
					num[A]++;
			}
		}

		int[] minCost = new int[10];
		for(int i = 0; i < 10; i++)
			minCost[i] = Int32.MaxValue;
		minCost[1] = 0;
		
		int[] perm = new int[10] {0,1,2,3,4,5,6,7,8,9};
		do
		{			
			int tmpCost = 0;
			for(int i = 0; i < 10; i++)
			{
				if(perm[i] == 1)
					break;
				
				tmpCost += c[perm[i],perm[i+1]];
			}

			minCost[perm[0]] = Math.Min(minCost[perm[0]], tmpCost);

		} while (nextPermutation(ref perm));
		
		int ans = 0;
		for(int i = 0; i < 10; i++)
			ans += num[i] * minCost[i];

		Console.WriteLine(ans);
	}


	private static bool nextPermutation(ref int[] array)
	{
		int len = array.Length;
		int i = len - 2;
		while((i >= 0) && (array[i] > array[i + 1]))
			i--;
		if(i < 0)
			return false;

		int pivot = array[i];
		int j = i + 1;
		while((j < len - 1) && (array[j + 1] > pivot))
			j++;
		
		int tmp = array[i];
		array[i] = array[j];
		array[j] = tmp;
		
		Array.Reverse(array, i + 1, len - i - 1);

		return true;
	}
}