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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import qualified Data.Map.Strict as M
import qualified Data.ByteString.Lazy.Char8 as B
import Data.Maybe
import Debug.Trace

qTrace :: (Show a) => String -> a -> a
-- qTrace lab a = trace (lab ++ ": " ++ show a) a
qTrace _ a = a

bsReadInts :: B.ByteString -> [Int]
bsReadInts = map (fst . fromJust . B.readInt) . B.words

type CMap = M.Map (Int, Int) Int

setMat :: [[Int]] -> CMap
setMat css =
  M.fromList $ concat (zipWith g [0..9] (map f css))
  where f cs = zip [0..9] cs
        g j zz = map (\(i, v) -> ((j, i), v)) zz

countAs :: [[Int]] -> M.Map Int Int
countAs ass = foldr f aInit [j | j <- concat ass, j >= 0]
  where f i mmap = M.alter g i mmap
        g Nothing = Just 1
        g (Just k) = Just (k + 1)
        aInit = M.fromList $ zip [0..9] (repeat 0)

modCmap :: CMap -> CMap
modCmap cmap = mcSub [(i,j) | i <- [0..9], j <- [0..9], i /= j] cmap

mcSub :: [(Int, Int)] -> CMap -> CMap
mcSub [] cmap = cmap
mcSub (p : tbc) cmap = mcSub ([q | q <- added, notElem q tbc] ++ tbc) newCmap
  where (added, newCmap) = mcSub2 p cmap

mcSub2 :: (Int, Int) -> CMap -> ([(Int, Int)], CMap)
mcSub2 (i,j) cmap = (map fst ps, foldr h cmap ps)
  where
    f k | i == k = []
        | j == k = []
        | otherwise = g i j k ++ g k i j
    g a b c | dist_abc < dist_ac = [((a,c), dist_abc)]
            | otherwise          = []
      where dist_abc = (cmap M.! (a,b)) + (cmap M.! (b,c))
            dist_ac  =  cmap M.! (a,c)
    ps = concatMap f [0..9]
    h ((a,b), dist) = M.adjust (const dist) (a,b)

solve :: CMap -> M.Map Int Int -> Int
solve cmap amap =
  let cmapMod = modCmap (qTrace "cmap" cmap)
  in sum $ map (\i -> (amap M.! i) * (cmapMod M.! (i, 1))) [0..9]

main :: IO ()
main = do
  conts <- B.getContents
  let [valH, valW] : iss = map bsReadInts (B.lines conts)
      (css, ass) = splitAt 10 iss
      cmat = setMat css
      xA = countAs ass
      answer = solve cmat xA
  putStrLn (show answer)