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::{Ordering};

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};

#[allow(unused_imports)]
use std::io::{stdin, stdout, BufWriter, Write};

#[allow(unused_imports)]
use std::iter::FromIterator;

#[allow(unused_macros)]
macro_rules! min {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::min($x, min!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! max {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::max($x, max!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! get { 
    ([$t: ty]) => { 
        { 
            let mut line = String::new(); 
            stdin().read_line(&mut line).unwrap(); 
            line.split_whitespace().map(|t|t.parse::<$t>().unwrap()).collect::<Vec<_>>()
        }
    };
    ([$t: ty];$n: expr) => {
        (0..$n).map(|_|get!([$t])).collect::<Vec<_>>()
    };
    ($t: ty) => {
        {
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t: ty),*) => {
        { 
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*)
        }
    };
    ($t: ty; $n: expr) => {
        (0..$n).map(|_|get!($t)).collect::<Vec<_>>()
    };
    ($($t: ty),*; $n: expr) => {
        (0..$n).map(|_|get!($($t),*)).collect::<Vec<_>>()
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

const BIG_STACK_SIZE: bool = true;

#[allow(dead_code)]
fn main() {
    use std::thread;
    if BIG_STACK_SIZE {
        thread::Builder::new()
            .stack_size(32 * 1024 * 1024)
            .name("solve".into())
            .spawn(solve)
            .unwrap()
            .join()
            .unwrap();
    } else {
        solve();
    }
}

pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T] where T: Ord {
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i-1]  {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i-1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j-1] < self[i-1]  {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i-1, j);

        true
    }

}

fn solve() {
    let (h,w) = get!(usize, usize);
    let mut c = get!([usize];10);
    let a = get!([isize];h);

    let mut nums = vec![0usize; 10];

    for row in a {
        for e in row {
            if e != -1 {
                nums[e as usize] += 1;
            }
        }
    }

    for k in 0..10 {
        for i in 0..10 {
            for j in 0..10 {
                c[i][j] = min!(c[i][j], c[i][k]+c[k][j]);
            }
        }
    }
    let mut sum = 0;
    for i in 0..10 {
        sum += nums[i]*c[i][1];
    }
    println!("{}", sum);
}