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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

program Project1;

{$APPTYPE CONSOLE}

uses
  SysUtils;
var
  h,w,i,j,count,x,y,sw,m,n:Integer;
  a,b:array of string;
  tf:Boolean;
begin
  try
    { TODO -oUser -cConsole Main : ここにコードを記述してください }
    Read(n);
    Readln(m);
    SetLength(a,n+1);
    SetLength(b,m+1);

    for i := 1 to n do
      Readln(a[i]);
    for i := 1 to m do
      Readln(b[i]);

    tf:=false;
    for j := 1 to n-m+1 do
      for i := 1 to n-m+1 do
        for y := 1 to m do begin
          for x := 1 to m do begin
            if b[y][x]<>a[j+y-1][i+x-1] then begin
              tf:=true;
              break;
            end;
            if (x=m)and(y=m) then begin
              Writeln('Yes');
              Readln;
              exit;
            end;
          end;
          if tf then begin
            tf:=false;
            break;
          end;
        end;

    Writeln('No');
    Readln;
  except
    on E: Exception do
      Writeln(E.ClassName, ': ', E.Message);
  end;
end.