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 (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use proconio::input;
use std::collections::HashMap;
use std::collections::VecDeque;
use std::cmp::max;
use std::cmp::min;


fn main() {
    input! {
        H: usize,
        W: usize,
    }

    let mut Cs: Vec<Vec<i64>> = Vec::new();

    for _i in 0..10 {
        input! {
            tmp: [i64; 10],
        }

        let mut arr: Vec<i64> = Vec::new();
        arr.extend_from_slice(&tmp);
        Cs.push(arr);
    }

    let mut As: Vec<Vec<i64>> = Vec::new();

    for _h in 0..H {
        input! {
           tmp: [i64; W],
        }

        let mut arr: Vec<i64> = Vec::new();
        arr.extend_from_slice(&tmp);
        As.push(arr);
    }

    // まずi => jの最小コストを求めておく
    // Cost[i][j][n] : i から j へ n回以下のステップで行った時の最小コスト
    let mut Cost: Vec<Vec<Vec<i64>>> = Vec::new();
    for _ in 0..10 {
        let mut arr2: Vec<Vec<i64>> = Vec::new();

        for _ in 0..10 {
            // コストの最大値は10^3を10回積み重ねた10^4
            // なので、それを上回る10^6としておこう
            let arr1 = vec![1000_000i64; 10];
            arr2.push(arr1);
        }

        Cost.push(arr2);
    }

    // まずn == 1 の場合
    for i in 0..10 {
        for j in 0..10 {
            Cost[i][j][1] = Cs[i][j];
        }
    }

    for n in 2..10 {
        for i in 0..10 {
            for j in 0..10 {
                // 直接行くか、kを経由して行くか

                let mut cand: i64 = Cs[i][j];

                for k in 0..10 {
                    cand = min(cand, Cs[i][k] + Cost[k][j][n - 1]);
                }

                Cost[i][j][n] = cand;
            }
        }
    }

    let mut ans = 0;

    for h in 0..H {
        for w in 0..W {
            if As[h][w] == 1 || As[h][w] == -1 {
                continue;
            }

            let a = &As[h][w];
            ans += Cost[*a as usize][1][9];
        }
    }

    println!("{}", ans);
}