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 DUMP(x)  cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define IFOR(i, m, n) for(int i = n - 1; i >= m; i-- )
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define FOREACH(x,a) for(auto& (x) : (a) )
#define ALL(v) (v).begin(), (v).end()
using namespace std;
typedef long long ll;
const ll INF = 1LL << 60;
/* テンプレートここまで */
typedef vector<vector<ll> > Matrix;

void warshall_floyd(Matrix& d) { // nは頂点数
  ll n = d.size();
  for (int i = 0; i < n; i++)      // 経由する頂点
    for (int j = 0; j < n; j++)    // 開始頂点
      for (int k = 0; k < n; k++)  // 終端
        d[j][k] = min(d[j][k], d[j][i] + d[i][k]);
}


int main() {
  ll H, W; cin >> H >> W;
  Matrix d = Matrix(10, vector<ll>(10, INF));
  REP(i, 10) REP(j, 10){
    int c;  cin >> c;
    d[i][j] = c;
  }
  vector<ll> cnt(10, 0);
  REP(i, H) REP(j, W){
    int a; cin >> a;
    if(a==-1) continue;
    cnt[a]++;
  }

  warshall_floyd(d);

  // REP(i, 10) REP(j, 10){
  //   printf("%lld % lld : %lld\n", i, j, d[i][j]);
  // }

  ll ans = 0;
  REP(i, 10){
    ans += cnt[i] * d[i][1];
  }
  cout << ans << endl;
}