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: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;


public class Main {

	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String input = br.readLine();
		String[] tmpArray = input.split(" ");
		int h = Integer.parseInt(tmpArray[0]);
		int w = Integer.parseInt(tmpArray[1]);

		int[][] costs = new int[10][10];

		for(int i = 0; i <= 9 ; i++){
			tmpArray = br.readLine().split(" ");
			for(int j = 0; j <= 9; j++){
				costs[i][j] = Integer.parseInt(tmpArray[j]);
			}
		}

		int[][] numbers = new int[h][w];
		for(int i = 0; i < h; i++){
			tmpArray = br.readLine().split(" ");
			for(int j = 0; j < w; j++){
				numbers[i][j] = Integer.parseInt(tmpArray[j]);
			}
		}
		
		int[][] minCosts = allPairsShortest(costs);
		
		long totalCost = 0;
		
		for(int i = 0; i < h; i++){
			for(int j = 0; j < w; j++){
				if(numbers[i][j] == -1 || numbers[i][j] == 0){
					continue;
				}
				
				else {
					totalCost += minCosts[numbers[i][j]][1];
				}
			}
		}
		
		System.out.println(totalCost);

	}

	static int[][] allPairsShortest(int[][] graph){
		int n = graph[0].length;
		int[][] dist = new int[n][n];
		
		for(int u = 0; u < n; u++){
			Arrays.fill(dist[u], Integer.MAX_VALUE);
			dist[u][u] = 0;
			
			for(int v = 0; v < n; v++){
				if(u != v){
					dist[u][v] = graph[u][v];
				}
			}
		}
		
		for(int k = 0; k < n; k++){
			for(int i = 0; i < n; i++){
				if(dist[i][k] == Integer.MAX_VALUE){
					continue;
				}
				
				for(int j = 0; j < n; j++){
					long newLen = dist[i][k];
					newLen += dist[k][j];
					
					if(newLen < dist[i][j]){
						dist[i][j] = (int)newLen;
					}
				}
			}
		}
		
		return dist;
	}
	
}