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;
use std::cmp;

// read a line(without line feed)
fn read_line() -> String {
    let mut in_str = String::new();
    io::stdin().read_line(&mut in_str).unwrap();
    return in_str.trim().to_string();
}

// read wods on a line
pub fn read_words_in_line() -> Vec<String> {
    let s = read_line();
    return s.split_whitespace().map(|x| x.to_string()).collect::<Vec<String>>()
}

// read type t on a line
pub fn read_t_in_line<T : std::str::FromStr>() -> Vec<T> {
    let mut v : Vec<T> = Vec::new();
    for e in read_words_in_line().iter().map(|x| x.parse::<T>()) {
        if let Ok(r) = e {
            v.push(r);
        }
        else{
            return Vec::new();
        }
    }
    return v;
}

// sort slice
pub fn sort<T : std::cmp::Ord>(slice : &mut [T]) {
    slice.sort();
}
// sort slice by reverse order
pub fn sort_rev<T : std::cmp::Ord>(slice : &mut [T]) {
    slice.sort_by(|x, y| y.cmp(x));
}

pub fn min<T : std::cmp::Ord>(a : T, b : T) -> T{
    return cmp::min(a, b);
}

pub fn max<T : std::cmp::Ord>(a : T, b : T) -> T{
    return cmp::max(a, b);
}



fn main() {

    let mut cs = vec![vec![0 ; 10] ; 10];

    let h_w = read_t_in_line::<i32>();
    let(h, w) = (h_w[0] as usize, h_w[1] as usize);

    let mut a = vec![vec![0 ; w] ; h];

    for i in 0..10 {
        let c_vec = read_t_in_line::<i32>();
        for j in 0..10 {
            cs[i][j] = c_vec[j];
        }
    }
    
    for i in 0..h {
        let a_vec = read_t_in_line::<i32>();
        for j in 0..w {
            a[i][j] = a_vec[j];
        }
    }

    // warshall floydでi -> jへ推移するコストの最小値を計算
    for k in 0..10 {
        for i in 0..10 {
            for j in 0..10 {
                cs[i][j] = min(cs[i][j], cs[i][k] + cs[k][j]);
            }
        }
    }

    let mut cost = 0;
    for i in 0..h {
        for j in 0..w {
            let t = a[i][j];
            if t != -1 {
                cost += cs[t as usize][1];
            }
        }
    }
    println!("{}", cost);

}