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 Lua (5.3.2) (judge verdict: Accepted)

local n, m = io.read("*n", "*n", "*l")
local a, b = {}, {}
for i = 1, n do
  local s = io.read()
  for j = 1, n do
    table.insert(a, s:sub(j, j))
  end
end
for i = 1, m do
  local s = io.read()
  for j = 1, m do
    table.insert(b, s:sub(j, j))
  end
end

local function check(x, y)
  for i = 1, m do
    for j = 1, m do
      local i_b = (i - 1) * m + j
      local i_a = (x + i - 2) * n + y + j - 1
      if b[i_b] ~= a[i_a] then return false end
    end
  end
  return true
end

local f = false
for i = 1, n + 1 - m do
  for j = 1, n + 1 - m do
    if check(i, j) then f = true break end
  end
  if f then break end
end
print(f and "Yes" or "No")