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 (1.3.71) (judge verdict: Accepted)

import java.io.BufferedReader
import java.io.InputStreamReader
import java.io.PrintWriter
import java.util.*
import kotlin.math.abs
import kotlin.math.cos
import kotlin.math.max
import kotlin.math.min

val br = BufferedReader(InputStreamReader(System.`in`))
val out = PrintWriter(System.out)

fun main() {
    br.use { out.use {
        solve()
        out.flush()
    } }
}

fun solve() {
    val (h, w) = readIntegerList()
    val map = Array(10) { readLongList().toLongArray() }
    warshallFloyd(map)
    val ans = Array(h) { readIntegerList() }
        .flatMap { it }
        .filter { it != -1 }
        .fold(0L) { acc, a -> acc + map[a][1] }
    out.println(ans)
}

/**
 * O((costMatrix.size)^3) costMatrix is changed
 * 全点対間最短路
 */
fun warshallFloyd(costMatrix: Array<LongArray>, LINF: Long = Long.MAX_VALUE) {
    val indices = costMatrix.indices
    for (k in indices)
        for (i in indices)
            for (j in indices)
                if (costMatrix[i][k] != LINF && costMatrix[k][j] != LINF)
                    costMatrix[i][j] = Math.min(costMatrix[i][j], costMatrix[i][k] + costMatrix[k][j])
}


fun readInteger() = Integer.parseInt(br.readLine())
fun readLong() = java.lang.Long.parseLong(br.readLine())
fun readStringList() = br.readLine()!!.split(' ')
fun readIntegerList() = readStringList().map(Integer::parseInt)
fun readLongList() = readStringList().map(java.lang.Long::parseLong)
fun readDoubleList() = readStringList().map(java.lang.Double::parseDouble)