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 Rust (1.42.0) (judge verdict: Accepted)

use std::io;
use std::str::FromStr;

fn input_vec() -> Vec<usize> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}

fn input_vec_i64() -> Vec<i64> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}


fn input_tuple() -> (u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

use std::cmp::Ordering;
use std::collections::BinaryHeap;
use std::usize;

#[derive(Copy, Clone, Eq, PartialEq)]
struct State {
    cost: usize,
    position: usize,
}

impl Ord for State {
    fn cmp(&self, other: &State) -> Ordering {
        other.cost.cmp(&self.cost)
            .then_with(|| self.position.cmp(&other.position))
    }
}

impl PartialOrd for State {
    fn partial_cmp(&self, other: &State) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

#[derive(Clone)]
struct Edge {
    node: usize,
    cost: usize,
}

// Dijkstra's shortest path algorithm.
fn shortest_path(adj_list: &Vec<Vec<Edge>>, start: usize) -> Vec<usize> {
    // dist[node] = current shortest distance from `start` to `node`
    let mut dist: Vec<usize> = (0..adj_list.len()).map(|_| usize::MAX).collect();

    let mut heap = BinaryHeap::new();

    dist[start] = 0;
    heap.push(State { cost: 0, position: start });

    while let Some(State { cost, position }) = heap.pop() {

        if cost > dist[position] { continue; }

        for edge in &adj_list[position] {
            let next = State { cost: cost + edge.cost, position: edge.node };

            if next.cost < dist[next.position] {
                heap.push(next);
                dist[next.position] = next.cost;
            }
        }
    }
    dist
}

fn main() {
    let (h, w) = input_tuple();
    let mut edges: Vec<Vec<Edge>> = vec![Vec::new(); 10];
    for i in 0..10 {
        let row = input_vec();
        for j in 0..10 {
            edges[j].push(Edge {node: i, cost: row[j]});
        }
    }
    let shortest_paths = shortest_path(&edges, 1);
    let mut sum = 0;
    for i in 0..h {
        let row = input_vec_i64();
        for j in row {
            if j != -1 {
                sum += shortest_paths[j as usize];
            }
        }
    }
    println!("{}", sum);
}