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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <functional>
#include <cassert>
#include <cctype>
#include <string>
#include <sstream>
#include <bitset>
using namespace std;
typedef long long int LL;
const int MAXN = 200 + 24;
int cost[10][10], num[MAXN][MAXN], dist[10], dis[10];
bool vis[10];
void dij(int st) {
	memset(vis, false, sizeof(vis));
	memset(dis, 0x3f3f3f3f, sizeof(dis));
	dis[st] = 0;
	for (int i = 0; i < 10; i++) {
		int u, mindis = 0x3f3f3f3f;
		for (int j = 0; j < 10; j++) {
			if (dis[j] < mindis && vis[j]==false) {
				mindis = dis[j]; u = j;
			}
		}
		vis[u] = true;
		for (int j = 0; j < 10; j++) {
			if (dis[j] > dis[u] + cost[u][j]) {
				dis[j] = dis[u] + cost[u][j];
			}
		}
	}
	dist[st] = dis[1];
}
int main() {
#ifdef kirito
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif // kirito
	int n,m;
	while (~scanf("%d%d", &n,&m)) {
		for (int i = 0; i < 10; i++) {
			for (int j = 0; j < 10; j++) {
				scanf("%d", &cost[i][j]);
			}
		}
		for (int i = 0; i < 10; i++) {
			dij(i);
		}
		LL res = 0;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				scanf("%d", &num[i][j]);
				if (num[i][j] != -1) {
					res += dist[num[i][j]];
				}
			}
		}
		cout << res << endl;
		
	}
	return 0;
}