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 Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

      program piyo
      integer,parameter :: p=8
      integer(p) :: n,m, i,j, offsetR,offsetC
      character(:),allocatable :: a,b
      integer(p),allocatable :: imA(:,:),imB(:,:),tempA(:,:)
      
      read*,n,m
      allocate( character(n)::a )
      allocate( character(m)::b )
      allocate( imA(N,N), imB(M,M), tempA(M,M) )
      do i = 1,n
        read*,a
        do j = 1,n
          if(a(j:j)=='.')then
            imA(i,j) = 1
          else
            imA(i,j) = 0
          end if
        end do
      end do
      do i = 1,m
        read*,b
        do j = 1,m
          if(b(j:j)=='.')then
            imB(i,j) = 1
          else
            imB(i,j) = 0
          end if
        end do
      end do
      
      do offsetR = 0,(N-M)
        do offsetC = 0,(N-M)
          tempA = imA(1+offsetR:M+offsetR,1+offsetC:M+offsetC)
          tempA = abs(tempA(:,:) - imB(:,:))
          if( sum(tempA(:,:))==0 )then
            print*,'Yes'
            stop
          end if
        end do
      end do
      
      print*,'No'
      
      stop
      
      do i = 1,n
        print'(*(i1,x))',ima(i,:)
      end do
      do i = 1,m
        print'(*(i1,x))',imb(i,:)
      end do
      
      
      
      end