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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#define M 200
#define N 10

int min(int c[])
{
	int min;
	if (c[0] == 0) min = c[1];
	else min = c[0];
	for (int i = 0; i <N; i++){
		if (c[i] < min&&c[i]!=0) min =c[i];
	}
	return min;
}
int minId(int c[])
{
	int min; int cnt;
	if (c[0] == 0) {
		min = c[1]; cnt = 1;
	}
	else {
		min = c[0]; cnt = 0;
	}
	for (int i = 0; i <N; i++){
		if (c[i] < min&&c[i] != 0) {
			cnt=i;
		}
	}
	return cnt;
}

void minAll(int c[][N], int re[],int id[])
{
	for (int i = 0; i < N; i++){
		re[i] = min(c[i]);
		id[i] = minId(c[i]);
	} 
}
int main(void)
{

	int H, W;
	int c[N][N];
	int A[M][M];
	int re[N], id[N];
	printf("H , W\n");
	scanf("%d", &H); scanf("%d", &W);
	printf("H=%d, W=%d\n", H, W);
	printf("cij\n");
	for (int i = 0; i < N; i++){
		printf("i=%d ", i);
		for (int j = 0; j < N; j++){
		 scanf("%d", &c[i][j]);
		}
	}
	printf("Aij\n");
	printf("H=%d, W=%d\n", H, W);
	for (int k = 0; k < H; k++){ 
		printf("i=%d\n", k);
		for (int l = 0; l < W; l++){
			scanf("%d", &A[k][l]);
		}
	}
	printf("Welcome!\n");
	minAll(c, re,id);
	//for (int i = 0; i < N; i++) printf("minc[%d]=%d\n", i, re[i]);
	//for (int i = 0; i < N; i++) printf("id c[%d]=%d\n", i, id[i]);
	puts(" ");
	int result = 0;
	for (int i = 0; i < H; i++){
		for (int j = 0; j < W; j++){
			if (A[i][j] == -1 || A[i][j] == 1) result += 0;
			else{
				while (A[i][j] != 1){
					result += re[A[i][j]];
					A[i][j] = id[A[i][j]];
				}
			}
		}
	}
	printf("%d\n", result);
	return 0;
}