Problem:
Score : 200 points

Problem Statement

You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.

A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.

Also, the given images are binary images, and the color of each pixel is either white or black.

In the input, every pixel is represented by a character: . corresponds to a white pixel, and # corresponds to a black pixel.

The image A is given as N strings A_1,...,A_N.

The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).

Similarly, the template image B is given as M strings B_1,...,B_M.

The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).

Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Constraints

1≦M≦N≦50

A_i is a string of length N consisting of # and ..

B_i is a string of length M consisting of # and ..

Input

The input is given from Standard Input in the following format:

N M
A_1
A_2
:
A_N
B_1
B_2
:
B_M

Output

Print Yes if the template image B is contained in the image A. Print No otherwise.

Sample Input 1

3 2
#.#
.#.
#.#
#.
.#

Sample Output 1

Yes

The template image B is identical to the upper-left 2 × 2 subimage and the lower-right 2 × 2 subimage of A. Thus, the output should be Yes.

Sample Input 2

4 1
....
....
....
....
#

Sample Output 2

No

The template image B, composed of a black pixel, is not contained in the image A composed of white pixels.

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Control.Monad
import Data.List
import Data.Bits
import Debug.Trace
import Data.Maybe
import qualified Data.ByteString.Char8 as C

main :: IO ()
main = do
  [n, m] <- readInts
  as <- getNLns n
  bs <- getNLns m
  putStrLn $ solve n m as bs

solve :: Int -> Int -> [String] -> [String] -> String
solve n m as bs =
  case foldM testMatch () genCandidates of
    Left  _ -> "Yes"
    Right _ -> "No"
  where
    d = n - m
    ds = [0 .. d]
    --
    testMatch _ (i, j) =
      let e = forM_ [0 .. m - 1] (\x ->
            if (drop j $ take (m + j) (as !! (i + x))) == bs !! x
              then Right ()
              else Left ()
            )
      in case e of
        Right _ -> Left ()
        Left  _ -> Right ()
    genCandidates :: [(Int, Int)]
    genCandidates = foldl' (\a x -> a ++ (zip (repeat x) ds)) [] ds


-- 以下ライブラリ

readInts :: IO [Int]
readInts = map (fst . fromJust . C.readInt) . C.words <$> C.getLine

readNInts :: Int -> IO [[Int]]
readNInts = flip replicateM readInts

readIntegers :: IO [Integer]
readIntegers = map (fst . fromJust . C.readInteger) . C.words <$> C.getLine

readNIntegers :: Int -> IO [[Integer]]
readNIntegers = flip replicateM readIntegers

getNLns :: Int -> IO [String]
getNLns = flip replicateM getLine

toTup2 :: [[a]] -> [(a, a)]
toTup2 = map (\ys -> case ys of y : z : zs -> (y, z))

encIdx2 :: Int -> (Int, Int) -> Int
encIdx2 mx (y, x) = y * mx + x

decIdx2 :: Int -> Int -> (Int, Int)
decIdx2 mx idx = (idx `div` mx, idx `mod` mx)

encIdx3 :: (Int, Int) -> (Int, Int, Int) -> Int
encIdx3 (my, mx) (z, y, x) = z * my * mx + y * mx + x

decIdx3 :: (Int, Int) -> Int -> (Int, Int, Int)
decIdx3 (my, mx) idx = (z, y, x)
  where
    basez = my * mx
    z = idx `div` basez
    y = (idx `mod` basez) `div` mx
    x = idx `mod` mx

while :: Monad m => m Bool -> m () -> m ()
while cond body = do
  b <- cond
  when b $ do
    body
    while cond body