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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"io"
	"math"
	"os"
	"strconv"
	"strings"

)


func main() {
	solve(os.Stdin, os.Stdout)
}

type ReaderEx struct {
	baseScanner *bufio.Scanner
	words       []string
	nextWordIdx func() int
	initWordIdx func(int)
}

func NewScannerEx(stdin io.Reader) *ReaderEx {
	var this ReaderEx
	var idx int
	var wordsCnt int
	this.baseScanner = bufio.NewScanner(stdin)
	this.baseScanner.Buffer(make([]byte, 0), 1E+11)
	this.nextWordIdx = func() int {
		cur := idx
		idx = (idx + 1) % wordsCnt
		return cur
	}
	this.initWordIdx = func(wc int) {
		idx = 0
		wordsCnt = wc
	}
	return &this
}

func (me *ReaderEx) Scan(doSplit bool) (isScanned bool) {
	isScanned = me.baseScanner.Scan()
	if doSplit {
		me.words = strings.Fields(me.baseScanner.Text())
	} else {
		me.words = []string{me.baseScanner.Text()}
	}
	me.initWordIdx(len(me.words))
	return
}

func (me *ReaderEx) nextBytes() []byte {
	return []byte(me.words[me.nextWordIdx()])
}
func (me *ReaderEx) nextString() string {
	return me.words[me.nextWordIdx()]
}

type StringArray []string

func (me *ReaderEx) nextStringArray() StringArray {
	return me.words[me.nextWordIdx():]
}

func (me StringArray) toStrings() []string {
	return me
}

func (me StringArray) toInts() []int {
	strValues := me.toStrings()
	values := make([]int, len(strValues))
	for i, v := range strValues {
		n, err := strconv.Atoi(v)
		if err != nil {
			panic(err)
		}
		values[i] = n
	}
	return values
}
func (me *ReaderEx) nextInt() int {
	n, err := strconv.Atoi(me.words[me.nextWordIdx()])

	defer func() {
		if err != nil {
			panic(fmt.Sprintf("getInt(%d)", n))
		}
	}()
	return n
}

func pow(x, y, mod int) (ans int) {

	ans = 1
	for bit := 1 << uint(math.Floor(math.Log2(float64(y)))); ; bit >>= 1 {
		if y&bit != 0 {
			ans *= x
			if mod > 0 {
				ans %= mod
			}
		}
		if bit <= 1 {
			return
		}
		ans *= ans
		if mod > 0 {
			ans %= mod
		}
	}
}

type factType struct {
	prime int
	count int
}

func fact(n int) (facts []*factType) {

	for i := 2; i*i <= n; i++ {
		if n%i == 0 {
			fa := factType{prime: i}
			for n%fa.prime == 0 {
				n /= fa.prime
				fa.count++
			}
			facts = append(facts, &fa)
		}
	}
	if n > 1 {
		facts = append(facts, &factType{prime: n, count: 1})
	}
	return facts
}
func dec2bin(n int) (ret []bool) {
	for bit := 1 << uint(math.Floor(math.Log2(float64(n)))); ; bit >>= 1 {
		ret = append(ret, n&bit != 0)

		if bit <= 1 {
			return
		}
	}

}

func abs(v int) int {
	if v < 0 {
		v = -v
	}
	return v
}

func b2i(b bool) (i int) {
	if b {
		i = 1
	} else {
		i = 0
	}
	return
}

func solve(stdin io.Reader, stdout io.Writer) {

	r := NewScannerEx(stdin)
	r.Scan(true)
	h, w := r.nextInt(), r.nextInt()

	cs := make([][]int, 10)
	for i := range cs {
		cs[i] = make([]int, 10)
	}
	for i := range cs {
		r.Scan(true)

		for j, v := range r.nextStringArray().toInts() {
			cs[i][j] = v
		}
	}

	as := make([][]int, h)
	for i := 0; i < h; i++ {
		r.Scan(true)
		as[i] = r.nextStringArray().toInts()
	}

	minInt := func(a int, vs ...int) int {
		for i := range vs {
			v := &vs[i]
			if a > *v {
				a = *v
			}
		}
		return a
	}
	for k := 0; k < 10; k++ {
		for i := 0; i < 10; i++ {
			for j := 0; j < 10; j++ {
				cs[i][j] = minInt(cs[i][j], cs[i][k]+cs[k][j])
			}
		}
	}

	result := 0
	for i := 0; i < h; i++ {
		for j := 0; j < w; j++ {
			if as[i][j] != -1 {
				result += cs[as[i][j]][1]
			}
		}
	}

	fmt.Fprintln(stdout, result)
}