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;
using ll = long long;

ll get_min_dist(ll start, ll goal, const vector<vector<ll>>& cs)
{
    vector<ll> dp(10, 1e9);
    vector<ll> dp_n(10, 1e9);

    dp[start] = 0;
    for (ll i = 0; i < 10; ++i) {
        for (ll j = 0; j < 10; ++j) {
            for (ll k = 0; k < 10; ++k) {
                dp_n[k] = min(dp_n[k], dp[j] + cs[j][k]);
            }
        }
        dp = move(dp_n);
        dp_n.assign(10, 1e9);
    }

    return dp[goal];
}

void main_()
{
    ll H, W;
    cin >> H >> W;
    vector<vector<ll>> cs(10, vector<ll>(10));
    for (auto&& c : cs) {
        copy_n(istream_iterator<ll>(cin), 10, begin(c));
    }
    vector<vector<ll>> As(H);
    for (ll i = 0; i < H; i++) {
        copy_n(istream_iterator<ll>(cin), W, back_inserter(As[i]));
    }

    vector<ll> min_dists_to_1(10);
    for (ll i = 0; i < 10; i++) {
        min_dists_to_1[i] = get_min_dist(i, 1, cs);
    }

    ll ans = 0;
    for (ll i = 0; i < H; i++) {
        ans += accumulate(
            begin(As[i]), end(As[i]),
            0LL, [&min_dists_to_1](auto acc, auto n) {
                return acc + ((n < 0) ? 0 : min_dists_to_1[n]);
            });
    }

    cout << ans << endl;
}
int main()
{
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);
    main_();
    return 0;
}