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 (1.15.1) (judge verdict: Accepted)

use std::io::stdin;
use std::time::SystemTime;

#[allow(dead_code)]
fn read_line() -> String {
    let mut s = String::new();
    stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

#[allow(dead_code)]
fn read<T>() -> T
where T: std::str::FromStr
{
    read_line().parse::<T>().ok().unwrap()
}

#[allow(dead_code)]
fn read_vec<T>() -> Vec<T>
where T: std::str::FromStr
{
    read_line().split(" ").map(|n| n.parse::<T>().ok().unwrap()).collect()
}

#[allow(dead_code)]
fn read_chars() -> Vec<char>
{
    read_line().chars().collect()
}

#[allow(dead_code)]
fn print_time() -> () {
    println!("{:?}", SystemTime::now());
}

#[allow(unused_variables)]
fn main() {
    let hw: Vec<i64> = read_vec();
    let mut maryoku: Vec<Vec<u64>> = vec![];
    for _ in 0..10 {
        maryoku.push(read_vec());
    }
    
    let mut map: Vec<Vec<i8>> = vec![];
    for _ in 0..hw[0] {
        map.push(read_vec());
    }
    
    for k in 0..10 {
        for i in 0..10 {
            for j in 0..10 {
                let t = maryoku[i][k] + maryoku[k][j];
                if t < maryoku[i][j] {
                    maryoku[i][j] = t;
                }
            }
        }
    }
    
    let mut count = 0;
    for i in 0..hw[0] {
        for j in 0..hw[1] {
            let tt = map[i as usize][j as usize];
            if tt != -1 && tt != 1 {
                let t = maryoku[tt as usize][1];
                count = count + t;
            }
        }
    }
    println!("{}", count);
}