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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n,m=gets.split.map &:to_i
a=n.times.map{gets.chomp.chars}
b=m.times.map{gets.chomp.chars}

#aの中で1つのマスに対してbを判定
ans=false
res=Array.new(m,false)
a.each_with_index do |aei,ai|
  a_break = false
  if (n-(ai+1)<m)
    break
  end
  aei.each_with_index do |aej,aj|
    if (n-(aj+1)<m)
      break
    else
      if aej === b[0][0]
        b.each_with_index do |bei,bi|
          b_break=false
          bei.each_with_index do |bej, bj|
			if a[ai+bi][aj+bj] === bej
              if (bj+1)==m
                res[bi]=true
              end
            end
          end
          break if b_break
        end
        break if a_break
      end
    end
  end
end
puts res.count(false)==0 ? 'Yes' : 'No'