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 Control.Monad
import Data.Maybe (fromJust)
import Debug.Trace

solve:: Int -> Int -> [String] -> [String] -> Bool
solve x y a b
    | find == True = True
    | y > maxS = False
    | x > maxS = solve 0 (y + 1) a b
    | otherwise = solve (x + 1) y a b
    where
        aLen = length a
        bLen = length b
        maxS = aLen - bLen
        subarea = (take bLen) . (drop maxS)
        aa = subarea (map subarea a)
        find = aa == b

main :: IO ()
main = do
    [n, m] <- map read . words <$> getLine :: IO [Int]
    a <- replicateM n getLine
    b <- replicateM m getLine
    putStrLn $ if (solve 0 0 a b) then "Yes" else "No"