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 ll long long
#define fw(p) for(int w=0;w<(p);w++)
#define fx(p) for(int x=0;x<(p);x++)
#define fy(p) for(int y=0;y<(p);y++)
#define fz(p) for(int z=0;z<(p);z++)
#define fyg(p,g) for(int y=(g);y<(p);y++)
#define fzg(p,g) for(int z=(g);z<(p);z++)
#define ce(d) cout<<d<<endl;
#define vecp(p) int aa;cin>>aa;(p).push_back(aa);
#define vecpl(p) long long aa;cin>>aa;(p).push_back(aa);
#define vecps(p) string aa;cin>>aa;(p).push_back(aa);
#define vecp2(p) cin>>aa;(p).push_back(aa);
#define vecpl2(p) long long a	b;cin>>ab;(p).push_back(ab);
#define vecps2(p) string ab;cin>>ab;(p).push_back(ab);
#define set0(k,n) for(int nn=0;nn<(n);nn++){ (k).push_back(0); }
#define sorts(c) sort((c).begin(),(c).end());
#define reverses(c) reverse((c).begin(),(c).end());
#define vec(b) vector<int> (b);
#define vecl(b) vector<long long> (b);
#define vecs(b) vector<string> (b);
#define vecsize(b,size) vector<int> (b)((size));
#define pb(b,a) (b).push_back((a));
#define doublece(a,b) cout<<(a)<<' '<<(b)<<endl;
#define pairs(s) vector<pair<int,int>> (s);
#define pairsl(s) vector<pair<ll,ll>> (s);
#define pairss(s) vector<pair<string,string>> (s);
#define pairsp(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairspl(s) ll aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsps(s) string aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsREV(s) (s).push_back(make_pair(bb,aa));
#define pairslREV(s) (s).push_back(make_pair(bb,aa));
#define pairssREV(s) (s).push_back(make_pair(bb,aa));
#define MOD 1000000007

int main() {
	
	int C[10][10];
	int H,W;
	cin>>H>>W;
	fx(10){
		fy(10){
			cin>>C[x][y];
		}
	}
	
	fx(10){
		fy(10){
			fz(10){
				C[y][z]=min(C[y][z],C[y][x]+C[x][z]);
			}
		}
	}
	int ans=0;
	fx(H){
		fy(W){
			int k;
			cin>>k;
			if(k!=-1){
				ans+=C[k][1];
			}
		}
	}
	ce(ans)
  
}