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: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"log"
	"os"
	"strconv"
)

// const abcd = "abcdefghijklmnopqrstuvwxyz"
// const ABCD = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

// var dy = [...]int{1, 1, 0, -1, -1, -1, 0, 1}
// var dx = [...]int{0, 1, 1, 1, 0, -1, -1, -1}

// var dx = [...]int{0, 1, 0, -1}
// var dy = [...]int{1, 0, -1, 0}

var inf int = 1e13
var mod = 1000000007
var next = newScanner()

// ---------------------------------------------------------

func main() {
	log.SetFlags(log.Lshortfile)
	H := next.Int()
	W := next.Int()
	c := make([][]int, 10)
	for i := 0; i < 10; i++ {
		c[i] = next.Ints(10)
	}
	A := make([]int, H*W)
	for i := 0; i < H*W; i++ {
		A[i] = next.Int()
	}

	for k := 0; k < 10; k++ {
		for i := 0; i < 10; i++ {
			for j := 0; j < 10; j++ {
				c[i][j] = min(c[i][j], c[i][k]+c[k][j])
			}
		}
	}
	for i := range c {
		fmt.Println(c[i])
	}
	ans := 0
	for i := range A {
		if A[i] != -1 && A[i] != 1 {
			// log.Println(i, A[i], c[1][A[i]])
			ans += c[A[i]][1]
		}
	}

	fmt.Println(ans)
}

// ---------------------------------------------------------

// Pair is...
type Pair struct {
	a, b int
}

// Pairs is sorted by []Pair struct
type Pairs []Pair

func (p Pairs) Len() int {
	return len(p)
}
func (p Pairs) Swap(i, j int) {
	p[i], p[j] = p[j], p[i]
}

func (p Pairs) Less(i, j int) bool {
	if p[i].a < p[j].a {
		return true
	} else if p[i].a == p[j].a {
		return p[i].b < p[j].b
	}
	return false
}

// ------int method-------------------------
func in(c, a, z int) bool {
	return c >= a && c < z
}
func btoi(b bool) int {
	if b {
		return 1
	}
	return 0
}
func itob(a int) bool {
	if a == 0 {
		return false
	}
	return true
}
func max(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r < a[i] {
			r = a[i]
		}
	}
	return r
}
func min(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r > a[i] {
			r = a[i]
		}
	}
	return r
}
func sum(a []int) int {
	r := 0
	for i := 0; i < len(a); i++ {
		r += a[i]
	}
	return r
}
func pro(a []int) int {
	r := 1
	for i := 0; i < len(a); i++ {
		r *= a[i]
	}
	return r
}
func fill(a []int, n int) []int {
	for i := 0; i < len(a); i++ {
		a[i] = n
	}
	return a
}
func minmax(a, b int) (int, int) {
	if a > b {
		return b, a
	}
	return a, b
}
func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

// ---------- buffered scanner -----------------------------------------
func iferr(err error) {
	if err != nil {
		panic(err)
	}
}

type scanner struct {
	r   *bufio.Reader
	buf []byte
	p   int
}

func newScanner() *scanner {
	rdr := bufio.NewReaderSize(os.Stdin, 10000)
	return &scanner{r: rdr}
}
func (s *scanner) next() string {
	s.pre()
	start := s.p
	for ; s.p < len(s.buf); s.p++ {
		if s.buf[s.p] == ' ' {
			break
		}
	}
	result := string(s.buf[start:s.p])
	s.p++
	return result
}
func (s *scanner) Line() string {
	s.pre()
	start := s.p
	s.p = len(s.buf)
	return string(s.buf[start:])
}
func (s *scanner) String() string {
	return s.next()
}
func (s *scanner) Int() int {
	v, e := strconv.Atoi(s.next())
	iferr(e)
	return v
}
func (s *scanner) Ints(n int) []int {
	r := make([]int, n)
	for i := 0; i < n; i++ {
		r[i] = s.Int()
	}
	return r
}
func (s *scanner) Int64() int64 {
	v, e := strconv.ParseInt(s.next(), 10, 64)
	iferr(e)
	return v
}
func (s *scanner) Uint64() uint64 {
	v, e := strconv.ParseUint(s.next(), 10, 64)
	iferr(e)
	return v
}
func (s *scanner) Float64() float64 {
	v, e := strconv.ParseFloat(s.next(), 64)
	iferr(e)
	return v
}
func (s *scanner) pre() {
	if s.p >= len(s.buf) {
		s.readLine()
		s.p = 0
	}
}
func (s *scanner) readLine() {
	s.buf = make([]byte, 0)
	for {
		l, p, e := s.r.ReadLine()
		iferr(e)
		s.buf = append(s.buf, l...)
		if !p {
			break
		}
	}
}