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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main(){
  #define int long long
  int H, W;
  cin >> H >> W;
  vector<vector<int>> vec(10, vector<int>(10));
  vector<vector<int>> tree(10, vector<int>(0));
  vector<int> mp(10);
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      if (i != j) {
        tree.at(i).push_back(j);
      }
    }
  }
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      cin >> vec.at(i).at(j);
    }
  }
  for (int i = 0; i <= 9; i++) {
    if (i == 1) {
      continue;
    }
    vector<int> dist(10,-1);
    queue<int> que;
    que.push(i);
    dist.at(i) = 0;
    while (!que.empty()) {
      int v = que.front();
      que.pop();
      if (v == 1) {
        continue;
      }
      for (int j = 0; j < (int) tree.at(v).size(); j++) {
        int nv = tree.at(v).at(j);
        if (dist.at(nv) != -1 && dist.at(nv) <= dist.at(v) + vec.at(v).at(nv)){
          continue;
        }
        //cout << v << " " << dist.at(v) << "   " << nv << " " << dist.at(nv) << endl;
        dist.at(nv) = dist.at(v) + vec.at(v).at(nv);
        que.push(nv);
      }
    }
    mp.at(i) = dist.at(1);
  }
  int answer = 0;
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      int a;
      cin >> a;
      if (a == -1) {
        continue;
      }
      answer += mp.at(a);
    } 
  }
  cout << answer << endl;
}