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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep2(i, x, n) for (int i = x; i < (n); i++)
#define all(n) begin(n), end(n)
struct cww
{
    cww()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
    }
} star;
const long long INF = numeric_limits<long long>::max();
typedef long long ll;
typedef vector<int> vint;
typedef vector<char> vchar;
typedef vector<vector<int>> vvint;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef unsigned long long ull;
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;
}
template <typename T>
vector<T> make_v(size_t a) { return vector<T>(a); }

template <typename T, typename... Ts>
auto make_v(size_t a, Ts... ts)
{
    return vector<decltype(make_v<T>(ts...))>(a, make_v<T>(ts...));
}
template <typename T, typename V>
typename enable_if<is_class<T>::value == 0>::type
fill_v(T &t, const V &v) { t = v; }

template <typename T, typename V>
typename enable_if<is_class<T>::value != 0>::type
fill_v(T &t, const V &v)
{
    for (auto &e : t)
        fill_v(e, v);
}
int main()
{
    int H, W;
    cin >> H >> W;
    auto c = make_v<ll>(10, 10);
    rep(i, 10)
    {
        rep(j, 10)
        {
            cin >> c[j][i];
        }
    }
    auto distance = c; //distance[i][j]は数字iを数字jに変えるときの最小コスト
    rep(i, 10)
    {
        rep(j, 10)
        {
            rep(k, 10)
            {
                chmin(distance[i][j], distance[i][k] + distance[k][j]);
            }
        }
    }

    ll ans = 0;
    rep(i, H)
    {
        rep(j, W)
        {
            int A;
            cin >> A;
            if (A != -1)
                ans += distance[A][1];
        }
    }
    cout << ans;
    return 0;
}