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 Python (2.7.6) (judge verdict: Accepted)


inp = []
inp.append(raw_input())
for i in xrange(10 + int(inp[0].split(" ")[0])):
    inp.append(raw_input())
Size = inp[0]

Mml = [None, 0] + [None] * 8
Cij = [[0] * 10 for i in range(10)]
for i in xrange(10):
    for j, a in enumerate(map(int,inp[i + 1].split(" "))):
        Cij[i][j] = a
        #print a,
    #print

def sloveMm(n):
    global Mml
    if(Mml[n] != None):
        return Mml[n]
    Nmm = Cij[n][1]
    rourt = [[n, 0, Nmm,n]]
    while len(rourt) > 0:
        rl = len(rourt)
        new_rourt = []
        for rn in xrange(rl):
            r = rourt[rn]
            for i in xrange(10):
                if i != r[0]:
                    nm = r[1] + Cij[r[0]][i]
                    if nm < Nmm:
                        """if Mml[i] != None:
                            Mml[n] = nm + Mml[i]
                            return nm + Mml[i]"""
                        if i == 1:
                            #print r
                            Nmm = nm
                        new_rourt.append([i, nm, nm + Cij[i][1]] + r[3:] + [i] )
        rourt = new_rourt
    Mml[n] = Nmm
    return Nmm


NumberList = [0] * 10
for Ai in inp[11:]:
    #print Ai
    for A in Ai.split(" "):
        if A != "-1" and A != "":
            NumberList[int(A)] += 1

ans = 0
for i, n in enumerate(NumberList):
    if n > 0:
        ans += sloveMm(i) * n
        #print "AA",i,":",Cij[i][1],">>",Mml[i],"*", n
#print Mml
#print NumberList
print ans