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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.io.BufferedReader
import java.io.InputStream
import java.io.InputStreamReader
import java.io.PrintWriter
import java.lang.StringBuilder
import java.util.*

fun PrintWriter.solve(sc: FastScanner) {
    val h = sc.nextInt()
    val w = sc.nextInt()
    val c = Array(10) { Array(10) { sc.nextInt() } }
    val a = Array(h) { Array(w) { sc.nextInt() } }
    val wf = Array(10) { Array(10) { 1000000000 } }
    for (i in 0 until 10) {
        for (j in 0 until 10) {
            if (i != j) {
                wf[i][j] = c[i][j]
            }
        }
    }
    for (k in 0 until 10) {
        for (i in 0 until 10) {
            for (j in 0 until 10) {
                if (wf[i][j] > wf[i][k] + wf[k][j]) {
                    wf[i][j] = wf[i][k] + wf[k][j]
                }
            }
        }
    }
    var ans = 0
    for (i in 0 until h) {
        for (j in 0 until w) {
            val v = a[i][j]
            if (v != -1 && v != 1) {
                ans += wf[v][1]
            }
        }
    }
    println(ans)
}

fun main(args: Array<String>) {
    val writer = PrintWriter(System.out, false)
    writer.solve(FastScanner(System.`in`))
    writer.flush()
}

class FastScanner(s: InputStream) {
    private var st = StringTokenizer("")
    private val br = BufferedReader(InputStreamReader(s))

    fun next(): String {
        while (!st.hasMoreTokens()) st = StringTokenizer(br.readLine())

        return st.nextToken()
    }

    fun nextInt() = next().toInt()
    fun nextLong() = next().toLong()
    fun nextLine() = br.readLine()
    fun nextDouble() = next().toDouble()
    fun ready() = br.ready()
}