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: Wrong Answer)

#![allow(unused_imports, unused_macros, dead_code)]
use std::f64::*;
use std::cmp::*;
use std::collections::*;

macro_rules! dump{
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

const INF: i64 = 0x3f3f3f3f3f3f3f3f;
const MOD: i64 = 1000000007;
const NOTHING: usize = std::usize::MAX;
const DX: [usize; 4] = [1, 0, std::usize::MAX, 0];
const DY: [usize; 4] = [0, 1, 0, std::usize::MAX];

fn main() {
    let (h, w): (usize, usize) = readln();
    let mut d: Vec<Vec<i64>> = Vec::with_capacity(10);
    let mut a: Vec<Vec<isize>> = Vec::with_capacity(h);
    for _ in 0..10 {
        let x: Vec<i64> = readln();
        d.push(x);
    }

    for _ in 0..h {
        let x: Vec<isize> = readln();
        a.push(x);
    }

    for i in 0..10 {
        for j in 0..10 {
            for k in 0..10 {
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    }

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

trait Read {
    fn read(s: &str) -> Self;
}

fn readln<T: Read>() -> T {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    T::read(buf.trim_right())
}

macro_rules! read_impl{
    ($($t:ty)*) => ($(
        impl Read for $t {
            fn read(s: &str) -> $t{
                s.parse().unwrap()
            }
        }
    )*)
}
read_impl! { usize u8 u16 u32 u64 isize i8 i16 i32 i64 f32 f64 }

impl Read for String {
    fn read(s: &str) -> Self {
        s.to_string()
    }
}

impl Read for Vec<char> {
    fn read(s: &str) -> Self {
        s.chars().collect()
    }
}

impl<T: Read> Read for Vec<T> {
    fn read(s: &str) -> Self {
        s.split_whitespace().map(T::read).collect()
    }
}

impl<A: Read, B: Read> Read for (A, B) {
    fn read(s: &str) -> Self {
        let tokens: Vec<_> = s.split_whitespace().collect();
        (A::read(tokens[0]), B::read(tokens[1]))
    }
}

impl<A: Read, B: Read, C: Read> Read for (A, B, C) {
    fn read(s: &str) -> Self {
        let tokens: Vec<_> = s.split_whitespace().collect();
        (A::read(tokens[0]), B::read(tokens[1]), C::read(tokens[2]))
    }
}

impl<A: Read, B: Read, C: Read, D: Read> Read for (A, B, C, D) {
    fn read(s: &str) -> Self {
        let tokens: Vec<_> = s.split_whitespace().collect();
        (A::read(tokens[0]), B::read(tokens[1]), C::read(tokens[2]), D::read(tokens[3]))
    }
}