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"
	"os"
	"strconv"
)

var (
	finv, inv, fac [MAX]int
)

func COM(n, k int) int {
	if n < k {
		return 0
	}
	if n < 0 || k < 0 {
		return 0
	}
	return fac[n] * (finv[k] * finv[n-k] % MOD) % MOD
}

func init() {
	fac[0] = 1
	fac[1] = 1
	finv[0] = 1
	finv[1] = 1
	inv[1] = 1
	for i := 2; i < MAX; i++ {
		fac[i] = fac[i-1] * i % MOD
		inv[i] = MOD - inv[MOD%i]*(MOD/i)%MOD
		finv[i] = finv[i-1] * inv[i] % MOD
	}
	sc.Split(bufio.ScanWords)
}

const (
	MOD = 1000000007
	MAX = 510000
)

func ModAdd(a, b int64) int64 {
	return (a + b) % MOD
}

func ModMul(a, b int64) int64 {
	return a * b % MOD
}

func ModSub(a, b int64) int64 {
	r := (a%MOD - b%MOD) % MOD
	if r < 0 {
		r += MOD
	}
	return r
}

func ModDiv(a, b int64) int64 {
	return a % MOD * ModInv(b) % MOD
}

func ModPow(a, n int64) int64 {
	var r int64 = 1
	for n > 0 {
		if n&1 == 1 {
			r = r * a % MOD
		}
		a = a * a % MOD
		n >>= 1
	}
	return r
}

func ModInv(a int64) int64 {
	var b, u, v int64 = MOD, 1, 0
	for b > 0 {
		t := a / b
		a -= t * b
		a, b = b, a
		u -= t * v
		u, v = v, u
	}
	u %= MOD
	if u < 0 {
		u += MOD
	}
	return u
}

type Q struct {
	value int
	index int
	A, B  int
}

type QS []*Q

func (p QS) Len() int           { return len(p) }
func (p QS) Less(i, j int) bool { return p[i].value > p[j].value }
func (pq QS) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

func (pq *QS) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Q)
	item.index = n
	*pq = append(*pq, item)
}

func (pq *QS) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	old[n-1] = nil  // avoid memory leak
	item.index = -1 // for safety
	*pq = old[0 : n-1]
	return item
}

func FactoredPrimeNumber(num int) map[int]int {
	m := map[int]int{}
	return factor(m, num, 2)
}

func factor(result map[int]int, num, pnum int) map[int]int {
	if pnum*pnum > num {
		if num != 1 {
			result[num] += 1
		}
		return result
	}

	if num%pnum == 0 {
		num /= pnum
		result[pnum]++
	} else {
		pnum++
	}
	return factor(result, num, pnum)
}

func GCD(a, b int64) int64 {
	if b == 0 {
		return a
	}
	return GCD(b, a%b)
}

func LCM(a, b int64) int64 {
	return a * b / GCD(a, b)
}

type UnionFind struct {
	par  []int
	size []int
}

func NewUnionFind(size int) *UnionFind {
	return new(UnionFind).init(size)
}

func (uf *UnionFind) init(size int) *UnionFind {
	uf.par = make([]int, size)
	uf.size = make([]int, size)

	for i := 0; i < size; i++ {
		uf.par[i] = i
		uf.size[i] = 1
	}

	return uf
}

func (uf *UnionFind) Unite(p int, q int) {
	rp := uf.Root(p)
	rq := uf.Root(q)
	if rp == rq {
		return
	}
	if uf.size[rp] < uf.size[rq] {
		uf.par[rp] = uf.par[rq]
		uf.size[rq] += uf.size[rp]
	} else {
		uf.par[rq] = uf.par[rp]
		uf.size[rp] += uf.size[rq]
	}
}

func (uf *UnionFind) Root(p int) int {
	for uf.par[p] == p {
		return p
	}
	uf.par[p] = uf.Root(uf.par[p])
	return uf.par[p]
}

func (uf *UnionFind) Size(p int) int {
	return uf.size[uf.Root(p)]
}

func (uf *UnionFind) Same(p int, q int) bool {
	return uf.Root(p) == uf.Root(q)
}

var sc = bufio.NewScanner(os.Stdin)

func getInt() int {
	n, _ := strconv.Atoi(getStr())
	return n
}

func getStr() string {
	sc.Scan()
	return sc.Text()
}

func main() {
	H, W := getInt(), getInt()
	C := [10][10]int{}
	for i := 0; i < 10; i++ {
		for j := 0; j < 10; j++ {
			C[i][j] = getInt()
		}
	}

	for i := 0; i < 10; i++ {
		for j := 0; j < 10; j++ {
			for k := 0; k < 10; k++ {
				if C[j][k] > C[j][i]+C[i][k] {
					C[j][k] = C[j][i] + C[i][k]
				}
			}
		}
	}
	ans := 0
	for i := 0; i < H; i++ {
		for j := 0; j < W; j++ {
			n := getInt()
			if n == -1 || n == 1 {
				continue
			}
			ans += C[n][1]
		}
	}
	fmt.Println(ans)
}

type IntHeap []int

func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IntHeap) Push(x interface{}) {
	// Push and Pop use pointer receivers because they modify the slice's length,
	// not just its contents.
	*h = append(*h, x.(int))
}

func (h *IntHeap) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}