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.PrintStream;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {
	static boolean debug = true;
	static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) {
		int k = sc.nextInt();
		int n = sc.nextInt();
		int[][] mp=new int[10][10];
		for(int i=0;i<10;i++){
			for(int j=0;j<10;j++){
				mp[i][j]=sc.nextInt();
			}
		}
		int[] mpCost=new int[10];
		for(int i=0;i<k*n;i++){
			int idx=sc.nextInt();
			if(idx>=0)
			mpCost[idx]+=1;
		}
		floyd(mp);
		long count=0;
		for(int i=0;i<mpCost.length;i++){
			count+=mpCost[i]*mp[i][1];
		}
		System.out.println(count);
		
		
	}

	private static void floyd(int[][] mp) {
		// TODO Auto-generated method stub
		for(int i=0;i<10;i++){
			for(int j=0;j<10;j++){
				for(int k=0;k<10;k++){
					mp[i][j]=Math.min(mp[i][j], mp[i][k]+mp[k][j]);
				}
			}
		}
	}

	private static void helper(int[] str) {
		// TODO Auto-generated method stub

	}

	private static void print(String string) {
		// TODO Auto-generated method stub
		if (debug)
			System.out.println(string);
	}

	private static void print(int[] data) {
		if (debug) {
			for (int i = 0; i < data.length; i++)
				System.out.println(i + ":" + data[i]);
		}
	}

	private static void print(String[] data) {
		if (debug) {
			for (int i = 0; i < data.length; i++)
				System.out.println(i + ":" + data[i]);
		}
	}

	private static void print(char[] data) {
		if (debug) {
			for (int i = 0; i < data.length; i++)
				System.out.println(i + ":" + data[i]);
		}
	}

	private static void print(double[] data) {
		if (debug) {
			for (int i = 0; i < data.length; i++)
				System.out.println(i + ":" + data[i]);
		}
	}

	private static void print(long[] data) {
		if (debug) {
			for (int i = 0; i < data.length; i++)
				System.out.println(i + ":" + data[i]);
		}
	}

	private static void print(int[][] data) {

		if (debug) {
			for (int i = 0; i < data.length; i++) {
				for (int j = 0; j < data[0].length; j++) {
					System.out.print(data[i][j] + " ");
				}
				System.out.println();
			}
		}
	}

	private static void print(long[][] data) {

		if (debug) {
			for (int i = 0; i < data.length; i++) {
				for (int j = 0; j < data[0].length; j++) {
					System.out.print(data[i][j] + " ");
				}
				System.out.println();
			}
		}
	}

	private static void print(char[][] data) {

		if (debug) {
			for (int i = 0; i < data.length; i++) {
				for (int j = 0; j < data[0].length; j++) {
					System.out.print(i + " " + j + ":" + data[i][j] + " ");
				}
				System.out.println();
			}
		}
	}

	private static void print(String[][] data) {

		if (debug) {
			for (int i = 0; i < data.length; i++) {
				for (int j = 0; j < data[0].length; j++) {
					System.out.print(i + " " + j + ":" + data[i][j] + " ");
				}
				System.out.println();
			}
		}
	}

	private static void print(double[][] data) {

		if (debug) {
			for (int i = 0; i < data[0].length; i++) {
				for (int j = 0; j < data.length; j++) {
					System.out.print(i + " " + j + ":" + data[i][j] + " ");
				}
				System.out.println();
			}
		}
	}

	public static void inPutData(char[][] c) {
		for (int i = 0; i < c.length; i++) {
			c[i] = sc.nextLine().toCharArray();
		}
	}

}