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 <bits/stdc++.h>
#define REP(i, a, n) for (int i = (a); i < (int)(n); ++i)
#define REPC(i, a, n) for (int i = (a); i <= (int)(n); ++i)
#define ALL(t) t.begin(), t.end()
#define RALL(t) t.rbegin(), t.rend()
#define MATINIT(type, row, col, init)                                          \
  vector<vector<type>>(row, vector<type>(col, init));
#define Yes(cond) cout << (cond ? "Yes" : "No") << endl;
#define YES(cond) cout << (cond ? "YES" : "NO") << endl;
using namespace std;
using LL = long long;
using ULL = unsigned long long;
template <class T> using VEC = std::vector<T>;
template <class T> using MAT = std::vector<std::vector<T>>;
void DUMP() { cerr << endl; }
template <class Head, class... Tail> void DUMP(Head &&head, Tail &&... tail) {
  cerr << head << ", ";
  DUMP(std::move(tail)...);
}
template <typename T> ostream &operator<<(ostream &os, vector<T> &vec) {
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, map<T1, T2> &m) {
  os << "{";
  for (auto p : m)
    os << p << ",";
  os << "}";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> p) {
  os << "[" << p.first << " " << p.second << "]";
  return os;
}

int main() {
  int H,W;
  cin>>H>>W;
  int n = 10;
  MAT<int> c=MATINIT(int,n,n,0);
  REP(i,0,n){
    REP(j,0,n){
      cin>>c[i][j];
    }
  }
  MAT<int> A=MATINIT(int,H,W,1);
  REP(i,0,H){
    REP(j,0,W){
      int t;
      cin>>t;
      A[i][j]=t==-1?1:t;
    }
  }
  REP(k,0,n){
    REP(i,0,n){
      REP(j,0,n){
        c[i][j]=min(c[i][j],c[i][k]+c[k][j]);
      }
    }
  }
  LL sum = 0;
  REP(i,0,H){
    REP(j,0,W){
      sum+=c[A[i][j]][1];
    }
  }
  cout<<sum<<endl;

  return 0;
}