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 <iomanip>
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <bitset>
#include <climits>
#define REP(i,n) for (int i=0;i<(n);i++)
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define RFOR(i,a,b) for (int i=(a)-1;i>=(b);i--)
#define ll long long
#define ull unsigned long long
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const int INF = 1e9;
const ll LL_INF = 1e18;
const int MOD = 1e9 + 7;

using namespace std;

struct Edge{
  int to, cost;
};
int dist[10][10];
  

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int H, W;
  cin >> H >> W;
  REP(i,10){
    REP(j,10){
      int c;
      cin >> c;
      if(c == 0) dist[i][j] = INF;
      else dist[i][j] = c;
    }
  }
  int ans = 0;
  REP(k,10){
    REP(i,10){
      REP(j,10){
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
      }
    }
  }
  REP(i,H){
    REP(j,W){
      int a;
      cin >> a;
      if(a == -1 or a == 1) continue;
      ans += dist[a][1];
    }
  }
  cout << ans << endl;
  return 0;
}