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(dead_code)]
fn getline() -> String {
    let mut res = String::new();
    std::io::stdin().read_line(&mut res).ok();
    res
}

#[allow(unused_macros)]
macro_rules! readl {
    ($t: ty) => {
        {
            let s = getline();
            s.trim().parse::<$t>().unwrap()
        }
    };
    ($( $t: ty),+ ) => {
        {
            let s = getline();
            let mut iter = s.trim().split(' ');
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*)
        }
    };
}

#[allow(unused_macros)]
macro_rules! readlvec {
    ($t: ty) => {
        {
            let s = getline();
            let iter = s.trim().split(' ');
            iter.map(|x| x.parse().unwrap()).collect::<Vec<$t>>()
        }
    }
}

#[allow(unused_macros)]
macro_rules! debug { ($x: expr) => { println!("{}: {:?}", stringify!($x), $x) } }
// macro_rules! debug { ($x: expr) => () }

#[allow(dead_code)]
fn show<T>(iter: T) -> String
where
    T: Iterator,
    T::Item: std::fmt::Display
{
    let mut res = iter.fold(String::new(), |sum, e| sum + &format!("{} ", e));
    res.pop();
    res
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BinaryHeap, HashMap, HashSet};
#[allow(unused_imports)]
use std::collections::btree_map::Entry::{Occupied, Vacant};

fn main() {
    use std::io::Write;
    let out = std::io::stdout();
    let mut out = std::io::BufWriter::new(out.lock());
    macro_rules! printf { ($($arg:tt)*) => (write!(out, $($arg)*).unwrap()); }
    macro_rules! printfln { () => (writeln!(out).unwrap()); ($($arg:tt)*) => (writeln!(out, $($arg)*).unwrap()); }
    let (h, w) = readl!(usize, usize);
    let mut c: Vec<_> = (0..10).map(|_| readlvec!(i32)).collect();
    for k in 0..10 {
        for i in 0..10 {
            for j in 0..10 {
                c[i][j] = min(c[i][k]+c[k][j], c[i][j]);
            }
        }
    }
    let a: Vec<_> = (0..h).map(|_| readlvec!(i32)).collect();
    let ans: i32 = a.into_iter().map(|row| {
        row.into_iter().filter(|&x| x != -1).map(|x| c[x as usize][1]).sum::<i32>()
    }).sum();
    printfln!("{}", ans);
}