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)

/*
* Author. Silviase(@silviasetitech)
* For AtCoder
*/

import java.util.*;
import java.lang.*;
import java.math.*;


public class Main {

    
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);

        int height = sc.nextInt();
        int width = sc.nextInt();
        int[][] needmagic = new int[10][10];
        int[][] wallnum = new int[height][width];
        int magicpoint = 0;
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < 10; j++) {
                needmagic[i][j] = sc.nextInt();
            }
        }
        for (int i = 0; i < height; i++) {
            for (int j = 0; j < width; j++) {
                wallnum[i][j] = sc.nextInt();
            }
        }

        for (int k = 0; k < 10; k++) {
            for (int i = 0; i < 10; i++) {
                for (int j = 0; j < 10; j++) {
                    needmagic[i][j] = Math.min(needmagic[i][j], needmagic[i][k] + needmagic[k][j]);
                }
            }
        }

        for (int i = 0; i < height; i++) {
            for (int j = 0; j < width; j++) {
                if (wallnum[i][j] == -1) {
                    continue;
                } else {
                    magicpoint += needmagic[wallnum[i][j]][1];
                }
            }
        }

        System.out.println(magicpoint);

        sc.close();
    }


    private static long gcd(Long m, long n) {
        if(m < n) {
            return gcd(n, m);
        }
        if(n == 0){
            return m;
        } 
        return gcd(n, m % n);
    }

    private static long[][] Combination_nCk(long n, long k) {
        n++; k++;
        long[][] ans = new long[(int)n][(int)k];
        for (int i = 0; i < n; i++) {
            ans[i][0] = 1;
        }
        for (int i = 0; i < n-1; i++) {
            for (int j = 0; j < k-1; j++) {
                if (i < j) {
                    ans[i][j] = 0;
                }else{
                    ans[i+1][j+1] = ans[i][j] + ans[i][j+1];
                }
            }
        }
        return ans;
    }
    private static double[][] Probability_nCk(int n, int k) {
        n++; k++;
        double[][] ans = new double[n][k];
        for (int i = 0; i < n; i++) {
            if (i == 0) {
                ans[i][0] = 1;
            }else{
                ans[i][0] = ans[i-1][0]/2;
            }
        }
        for (int i = 0; i < n-1; i++) {
            for (int j = 0; j < k-1; j++) {
                if (i < j) {
                    ans[i][j] = 0;
                }else{
                    ans[i+1][j+1] = (ans[i][j] + ans[i][j+1])/2;
                }
            }
        }
        return ans;
    }
}