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 Python (3.4.3) (judge verdict: Runtime Error)

n, m = [int(x) for x in input().split()]
img_a = []
img_b = []

for i in range(n):
    img_a.append(input())

for i in range(m):
    img_b.append(input())

ans = "No"
for a in range(n - m + 1):
    dup_a = img_a.copy()
    dup_b = img_b.copy()
    while True:
        for b in range(m):
            if dup_b[b] in dup_a[b+a]:
                if b == 0:
                    index = dup_a[a].index(dup_b[0])
                current_index = dup_a[b+a].index(dup_b[b])
                
                if current_index > index:
                    for line in dup_a:
                        del line[:current_index]
                        break
                elif current_index < index:
                    for line in dup_a:
                        del line[:index]
                        break
        
            else:   # if not substring in string.
                break
        else: # if not loop broken.
            ans = "Yes"
            break
        if not dup_b[b] in dup_a[b+a]:
            break

print(ans)