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: Wrong Answer)

import Data.List

main = do
  li <- getLine
  let [n,m] = map read $ words li
  co <- getContents
  let (as,bs) = splitAt n $ lines co
  let ans = compute n m as bs
  putStrLn $ if ans then "Yes" else "No"

compute :: Int -> Int -> [String] -> [String] -> Bool
-- compute n m as bs = any (bs ==) $ concatMap (map (cut n m)) $ cut n m as
compute n m as bs = not $ null
  [ ()
  | as1 <- cut n m as
  , bs1 <- map (cut n m) as1
  , bs1 == bs ]

-- cut n m xs : 長さnのリストxsから、長さmの部分リストを全て切り出す
cut :: Int -> Int -> [a] -> [[a]]
cut n m xs = map (take m) $ take (n-m+1) $ tails xs