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::collections::HashMap;
//use std::collections::BinaryHeap;
//use std::collections::HashSet;
//use std::mem; // mem::swap(&mut x,&mut y);
use std::cmp; // cmp::max,cmp::min

// Rustメモ: https://hackmd.io/i0lQY1OKTDW1t66TECOmiw?both

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

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

#[allow(dead_code)]
fn read_ints() -> Vec<i64>{
    let s = read_line();
    let split:Vec<&str> = s.split(" ").collect();
    split.iter().map(|&x| x.to_string().parse().unwrap()).collect()
}

#[allow(dead_code)]
fn read_int() -> i64{
    read_ints()[0]
}

#[allow(dead_code)]
fn reverse(s: &String) -> String{
    s.chars().rev().collect::<String>()
}

#[allow(dead_code)]
fn read_lines(n:i64) -> Vec<String>{
    let mut xs = Vec::new();

    for _i in 0..n{
        xs.push(read_line());
    }

    xs
}

#[allow(dead_code)]
fn read_ints2(n:i64) -> Vec<i64>{
    read_lines(n).iter().map(|x| x.parse().unwrap()).collect()
}

////////////////////////////////////////////////

fn calc_min_cost(cost:&mut Vec<Vec<i64>>){
    for _ in 0..10{
        for i0 in 0..10{
            for i1 in 0..10{
                for i2 in 0..10{
                    cost[i0][i2] = cmp::min(
                        cost[i0][i2],
                        cost[i0][i1]+cost[i1][i2]);
                }
            }
        }
    }
}

fn main(){
    let mut cost = Vec::new();
    let mut board = Vec::new();
    let input = read_ints();
    let (h,w) = (input[0],input[1]);
    let mut ans = 0;

    for _ in 0..10{
        cost.push(read_ints());
    }
    for _ in 0..h{
        board.push(read_ints());
    }
    calc_min_cost(&mut cost);

    for i in 0..h as usize{
        for j in 0..w as usize{
            if board[i][j] != -1 {
                ans += cost[board[i][j] as usize][1];
            }
        }
    }
    println!("{}",ans);
}