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 ALL(x) (x).begin(), (x).end()
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define pb push_back
#define SZ(x) ((int)((x).size()))

typedef long long ll;

const ll mod = 1000000007;
const double pi = 3.1415926535897932384626;

ll powmod(ll a, ll b) {ll res = 1; ll rem = a % mod; for (; b; b >>= 1) {if (b & 1) res = (res * rem) % mod; rem = (rem * rem) % mod;} return res;}

const int N = 18;

ll G[N][N];
ll A[N];
ll dis[N];

ll gao(int s)
{
	memset(dis, -1, sizeof(dis));
	priority_queue< pair<ll, int>, vector< pair<ll, int> >, greater< pair<ll, int> > > Q;
	Q.push(mp(0ll, s));
	dis[s] = 0ll;

	while (!Q.empty()) {
		ll d = Q.top().fi;
		int u = Q.top().se;
		dis[u] = d;

		Q.pop();

		if (u == 1) return dis[u];

		for (int i = 0; i <= 9; ++i) {
			int v = i;
			ll _d = d + G[u][v];			
			if (dis[v] == -1 || _d < dis[v]) {
				dis[v] = _d;
				Q.push(mp(_d, v));
			}
		}
	}

	return 0ll;
}		

int main()
{	
#ifdef hannibal_08
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
#endif
    
	int n = 0;
	int m = 0;
	scanf("%d %d", &n, &m);

	for (int i = 0; i <= 9; ++i) {
		for (int j = 0; j <= 9; ++j) {
			scanf("%lld", &G[i][j]);
		}
	}

	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			ll x = 0ll;
			scanf("%lld", &x);
			if (x != -1 && x != 1) ++A[x];
		}
	}

	ll ret = 0ll;

	for (int i = 0; i <= 9; ++i) {
		if (i == 1) continue;
		ll ss = gao(i);
		ret += (ss * A[i]);
	}

	printf("%lld\n", ret);
	
	return 0;
}