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;
#define ALL(x) (x).begin(),(x).end()
#define COUT(x) cout<<(x)<<"\n"
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define REP(i, n) for(int i=0;i<n;i++)
#define YES(x) cout<<(x?"YES":"NO")<<"\n"
#define Yes(x) cout<<(x?"Yes":"No")<<"\n"
#define dump(x) cout<<#x<<" = "<<(x)<<"\n"
#define endl "\n"
using G = vector<vector<int>>;
using M = map<int,int>;
using P = pair<int,int>;
using PQ = priority_queue<int>;
using PQG = priority_queue<int,vector<int>,greater<int>>;
using V = vector<int>;
using ll = long long;
using edge = struct { int to; int cost; };
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

vector<int> d(10, INF);
vector<edge> g[10];

void dijkstra(int s) {
  priority_queue<P, vector<P>, greater<P>> que;
  fill(d.begin(), d.end(), INF);
  d[s] = 0;
  que.push(P(0, s));

  while (!que.empty()) {
    P p = que.top(); que.pop();
    int v = p.second;
    if (d[v] < p.first) continue;

    REP(i,g[v].size()) {
      edge e = g[v][i];
      if (d[e.to] > d[v] + e.cost) {
        d[e.to] = d[v] + e.cost;
        que.push(P(d[e.to], e.to));
      }
    }
  }
}

int main() {
  IOS;
  int h, w; cin >> h >> w;
  REP(i,10) {
    REP(j,10) {
      int cost; cin >> cost;
      g[i].push_back({ j, cost });
    }
  }

  int memo[10];
  REP(i,10) {
    dijkstra(i);
    memo[i] = d[1];
  }

  int ans = 0;
  REP(i,h) REP(j,w) {
    int a; cin >> a;
    if (a == -1) continue;
    ans += memo[a];
  }

  COUT(ans);
  return 0;
}