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 {
	static int[][] c = new int[10][10];

	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int h = sc.nextInt();
		int w = sc.nextInt();
		for (int i = 0; i < 10; i++) {
			for (int j = 0; j < 10; j++) {
				c[i][j] = sc.nextInt();
			}
		}
		int[] arr = new int[10];
		for (int i = 0; i < 10; i++) {
			if (i == 1) {
				continue;
			}
			int ceiling = c[i][1];
			for (int j = 0; j < 10; j++) {
				if (j == i || j == 1) {
					continue;
				}
				int ret = costCheck(j, c[i][j], ceiling);
				if (ret < ceiling) {
					ceiling = ret;
				}
			}
			arr[i] = ceiling;
		}
		int total = 0;
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				int x = sc.nextInt();
				if (x != -1) {
					total += arr[x];
				}
			}
		}
		System.out.println(total);
	}
	
	public static int costCheck(int select, int cost, int ceiling) {
		int direct = c[select][1] + cost;
		for (int i = 0; i <= 9; i++) {
			if (i != 1 && i != select) {
				if (direct > c[select][i] + cost && cost + c[select][i] < ceiling) {
					int ret = costCheck(i, cost + c[select][i], ceiling);
					if (ret < direct) {
						direct = ret;
					}
				}
			}
		}
		if (direct < ceiling) {
			return direct;
		} else {
			return ceiling;
		}
	}
}