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 Common Lisp (SBCL 1.1.14) (judge verdict: Runtime Error)

#include<iostream>
#include<string.h>

using namespace std;
const int inf = 1e7;
const int man = 1e2+10;
int maze[man][man];
int vis[man],dis[man];
int c[15][15];
int jishu[10];
int h,w;

int dij(int u)
{
	memset(vis,0,sizeof(vis));
	for(int i =1 ;i <= 9 ;i++)
	{
		dis[i] = c[u][i];
	}
	vis[u] = 1;
	for(int i =1 ;i < 9;i++)//必须是小于9，不能等于，debug了好久。。。
	{
		int minn = inf ,v = -1;
		for(int j =1 ;j <= 9 ;j++)
		{
			if(!vis[j]&&dis[j]<minn)
			{
				v = j;
				minn = dis[j];
			}
		}
		vis[v] = 1;
		for(int j = 1; j <= 9 ;j++)
		{
			dis[j] = min(dis[j],dis[v]+c[v][j]);
		}
	}
	return dis[1];
}

void solve()
{
	while(cin >> h >> w)
	{
		for(int i = 0; i < 10 ;i++)
		{
			for(int j = 0 ;j < 10 ;j++)
			{
				cin >> c[i][j];
			}
		}
		memset(jishu,0,sizeof(jishu));
		for(int i = 1 ; i <= w ;i++)
		{
			for(int j = 1; j<= h ;j++)
			{
				cin >> maze[i][j];
				if(maze[i][j] > 1)
				{
					jishu[maze[i][j]]++;
				}
			}
		}
		int ans = 0;
		//cout << dij(8) << endl;
		for(int i = 1 ; i <= 9 ;i++)
		{
			if(jishu[i])
			{
				int k = jishu[i];
				ans= ans +dij(i)*k;
			}
		//	cout << jishu[i] << endl;
		}
		cout << ans << endl;
	}
	
}

int main()
{
	solve();
	return 0;
}