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<algorithm>
#include<cmath>
#include<iomanip>
#include<set>
#include<map>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
const double pi=3.14159265358979323846;
const int inf=1e9;
const ll INF=1e18;
typedef pair<int,int> P;
int main() {
  cin.tie(0),cout.tie(0);
  ios::sync_with_stdio(false);
  ll h,w,a[205][205]={},c[10][10]={},ans=0;;
  cin >> h >> w;
  for(int i=0; i<10; i++) {
    for(int j=0; j<10; j++) {
      cin >> c[i][j];
    }
  }
  for(int i=0; i<h; i++) {
    for(int j=0; j<w; j++) {
      cin >> a[i][j];
    }
  }
  for(int i=0; i<10; i++) {
    for(int j=0; j<10; j++) {
      for(int k=0; k<10; k++) {
        if(i!=j&&i!=k&&j!=k) {
          c[j][k]=min(c[j][k],c[j][i]+c[i][k]);
        }
      }
    }
  }
  for(int i=0; i<h; i++) {
    for(int j=0; j<w; j++) {
      if(a[i][j]!=1&&a[i][j]!=-1) {
        ans+=c[a[i][j]][1];
      }
    }
  }
  cout << ans << endl;
}