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>
using namespace std;

#define each(i, c) for (auto& i : c)
#define mkp(a, b) make_pair(a, b)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> Pll;
const ll MOD = 1e9+7;

template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p) { os << "(" << p.first << ": " << p.second << ")"; return os; }
template<typename T> ostream& operator << (ostream& os, vector<T> v) { os << "("; each (i, v) os << i << ", "; os << ")"; return os; }
template<typename K, typename V> ostream& operator << (ostream& os, map<K, V> m) { os << "{"; each (i, m) os << i << ", "; os << "}"; return os; }

typedef struct edge {
  ll to;
  ll cost;
} edge;
typedef vector<vector<edge>> G; // graph
typedef vector<Pll> GR; // graph result (cost, vertex num)

GR dijkstra(G &g, ll s) { // O(ElogV)
  priority_queue<Pll, vector<Pll>, greater<Pll>> q;
  GR res(g.size(), mkp(1e18, 0));
  
  res[s] = mkp(0, s);
  q.push(res[s]);

  while (!q.empty()) {
    auto p = q.top();
    q.pop();

    ll cost = p.first;
    ll pos = p.second;
    if (res[pos].first < cost) continue;

    each (j, g[pos]) {
      ll to = j.to;
      ll to_cost = j.cost + cost;
      if (to_cost >= res[to].first) continue;
      res[to].first = to_cost;
      q.push(mkp(to_cost, to));
    }
  }
  return res;
}

int main() {
  ll h, w;
  cin >> h >> w;

  G g(10);
  for (ll i = 0; i < 10; ++i) {
    for (ll j = 0; j < 10; ++j) {
      ll cost;
      cin >> cost;
      g[i].push_back(edge{.to=j, .cost=cost});
    }
  }

  ll ans = 0;
  for (ll i = 0; i < h; ++i) {
    for (ll i = 0; i < w; ++i) {
      ll n;
      cin >> n;
      if (n == -1) continue;
      if (n == 1) continue;
      auto res = dijkstra(g, n);
      ans += res[1].first;
    }
  }

  cout << ans << endl;
  
  return 0;
}