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 <vector>
#include <stack>
#include <queue>
#include <string>
#include <algorithm>
#include <math.h>
#include <map>
using namespace std;
#define lint long long
#define ll long long
static const int INF = 1 << 22;
#define pii pair<long long,long long>
#define all(x) (x).begin(),(x).end()
#define mod 1000000007


#define MAX_V 10; 
int N = 10; // N = 10(const.)
ll G[10][10]; ll A[210][210];
ll d[10];
bool used[10];
ll mt0[10];
void dijkstra(int s) {
	for (int i = 0; i < 10; i++) {
		d[i] = INF;
		used[i] = false;
	}
	d[s] = 0;

	while (true) {
		int v = -1;
		for (int i = 0; i < N; i++) {
			if (!(used[i]) && (v == -1 || d[v] > d[i])) v = i;
		}
		if (v == -1) break;
		used[v] = true;

		for (int i = 0; i < N; i++) {
			d[i] = min(d[i], d[v] + G[v][i]);
		}
	}
}
int main(void) {
	long long H; cin >> H;
	long long W; cin >> W;

	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			cin >> G[i][j];
		}
	}

	for (int i = 0; i < N; i++) {
		dijkstra(i);
		mt0[i] = d[1];
	}

	ll ans = 0;

	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			cin >> A[i][j];
			if (A[i][j] != -1) {
				ans += mt0[A[i][j]];
			}
		}
	}

	cout << ans << endl;

	return 0;
}