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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

void warshall_fl(int* c){
  int tmp;
  for (int i = 0; i < 10; i++){
    for (int j = 0; j < 10; j++){
      //if(i == j) continue;
      for(int k = 0; k < 10; k++){
        tmp = c[i*10+k] + c[k*10+j];
        if(c[i*10+j] > tmp) c[i*10+j] = tmp;
      }
    }
  }
}

int main (void) {
  int h, w;
  int c[10*10] = {0};
  cin >> h >> w;
  cin.ignore();
  for (int i = 0; i < 10; i++){
    for (int j = 0; j < 10; j++){
      cin >> c[i*10+j];
    }
    cin.ignore();
  }
  warshall_fl(c);
  int tmp;
  unsigned long long sum = 0;
  int a[10] = {0};
  for (int i = 0; i < h; i++){
    for(int j = 0; j < w; j++){
      cin >> tmp;
      if(tmp!=-1) {
        sum += c[tmp*10+1];
        a[tmp]++;
      }
    }
    cin.ignore();
  }

  cout << sum << endl;
/*
  for (int i = 0; i < 10; i++) cout << a[i] << "," << flush;
  cout << endl;
  for (int i = 0; i < 100; i++){
    cout << c[i] << "," << flush;
    if(i != 0 && i%10 == 9) cout << c[i] << endl;
  }
*/
  sum = 0;
  for(int i = 0; i < 10; i++){
    if(i == 1) continue;
    if(a[i] >= 0) sum += a[i] * c[i*10+1];
  }
  //cout << sum << endl;
  return 0;
}