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)

// =========
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet};
#[allow(unused_imports)]
use std::process::exit;

#[allow(dead_code)]
const MOD: usize = 1000000007;
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    // var... 変数の識別子, $t...型を一つよむ
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        //ここで繰り返し
        input_inner!{$iter $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    //
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    // 配列の最後のNestではここで型が指定されてparseされる
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}
// =========
fn warshall_floyd(adjm: &mut Vec<Vec<Option<usize>>>) {
    let n = adjm.len();
    // kを経て短くなる辺を全部短くする
    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                if let (Some(dik), Some(djk)) = (adjm[i][k], adjm[k][j]) {
                    if adjm[i][j].is_none() || adjm[i][j].unwrap() > dik + djk {
                        adjm[i][j] = Some(dik + djk);
                    }
                }
            }
        }
    }
}
fn main() {
    input! {
        h: usize,
        w: usize,
        costs: [[usize;10];10],
        wall: [[i64;w];h]
    }
    let mut adjm: Vec<Vec<Option<usize>>> = vec![vec![None; 10]; 10];
    for i in 0..10 {
        for j in 0..10 {
            adjm[i][j] = Some(costs[i][j]);
        }
    }
    warshall_floyd(&mut adjm);
    // iを家として、最長の時間が最も短いもの
    let mut ans = 0;
    for i in 0..h {
        for j in 0..w {
            if wall[i][j] != -1 {
                ans += adjm[wall[i][j] as usize][1].unwrap();
            }
        }
    }
    println!("{:?}", ans);
}