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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cassert>
#include <tuple>
#include <array>

using namespace std;

constexpr long long mod = static_cast<long long>(1e9 + 7);

int main() {
    int H, W;
    cin >> H >> W;

    int path[10][10];

    for(int i = 0; i < 10; ++i){
        for(int j = 0; j < 10; ++j){
            cin >> path[i][j];
        }
    }

    vector<vector<long long>> wall(H, vector<long long>(W, 0));

    for(auto& vec: wall) {
        for(auto& val: vec){
            cin >> val;
        }
    }

    long long answers[10] = {};

    for(int i = 0; i < 10; ++i) {
        auto comp = [](auto x, auto y){return std::get<0>(x) > std::get<0>(y);};
        int costs[10];
        fill(costs, costs + 10, -1);
        priority_queue<tuple<long long, int>, vector<tuple<long long, int>>, decltype(comp)> pque(comp);
        pque.push(make_tuple(0, i));

        while(!pque.empty()) {
            auto state = pque.top();
            pque.pop();
            auto cost = std::get<0>(state);
            auto pos = std::get<1>(state);

            costs[pos] = cost;
            if(pos == 1) break;
            for(int j = 0; j < 10; ++j) {
                if(costs[j] < 0 || cost + path[pos][j] < costs[j]) {
                    pque.push(make_tuple(cost + path[pos][j], j));
                }
            }
        }
        answers[i] = costs[1];
    }

    long long ans = 0;
    for(auto& vec: wall){
        for(auto val: vec) {
            if(val > 0) ans += answers[val];
        }
    }

    cout << ans << endl;

    return 0;
}