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 (3.4.3) (judge verdict: Accepted)

import sys
## io ##
def IS(): return sys.stdin.readline().rstrip()
def II(): return int(IS())
def MII(): return list(map(int, IS().split()))
def MIIZ(): return list(map(lambda x: x-1, MII()))
## dp ##
def DD2(d1,d2,init=0): return [[init]*d2 for _ in range(d1)]
def DD3(d1,d2,d3,init=0): return [DD2(d2,d3,init) for _ in range(d1)]
## math ##
def to_bin(x: int) -> str: return format(x, 'b') # rev => int(res, 2)
def to_oct(x: int) -> str: return format(x, 'o') # rev => int(res, 8)
def to_hex(x: int) -> str: return format(x, 'x') # rev => int(res, 16)
MOD=10**9+7
def divc(x,y) -> int: return -(-x//y)
def divf(x,y) -> int: return x//y
def gcd(x,y):
    while y: x,y = y,x%y
    return x
def lcm(x,y): return x*y//gcd(x,y)
def enumerate_divs(n):
    """Return a tuple list of divisor of n"""
    return [(i,n//i) for i in range(1,int(n**0.5)+1) if n%i==0]
def get_primes(MAX_NUM=10**3):
    """Return a list of prime numbers n or less"""
    is_prime = [True]*(MAX_NUM+1)
    is_prime[0] = is_prime[1] = False
    for i in range(2, int(MAX_NUM**0.5)+1):
        if not is_prime[i]: continue
        for j in range(i*2, MAX_NUM+1, i):
            is_prime[j] = False
    return [i for i in range(MAX_NUM+1) if is_prime[i]]
## libs ##
from itertools import accumulate as acc
from collections import deque, Counter
from heapq import heapify, heappop, heappush
from bisect import bisect_left

#======================================================#

def main():
    h,w = MII()
    d = [MII() for i in range(10)]
    wall = []
    for i in range(h):
        wall += MII()
    for k in range(10):
        for i in range(10):
            for j in range(10):
                d[i][j] = min(d[i][j], d[i][k]+d[k][j])
    ans = sum([d[w][1] for w in wall if w != -1])
    print(ans)

if __name__ == '__main__':
    main()