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 (gfortran v4.8.4) (judge verdict: Accepted)

program abc054b
    use,intrinsic :: iso_fortran_env
    implicit none
    integer(int32):: n,m,i,j
    character(1),allocatable:: a(:,:), b(:,:)

    read*, n,m
    allocate(a(n,n),b(m,m))
    input_char : block
        character(n):: inpt_a
        character(m):: inpt_b
        do i=1,n
            read*, inpt_a
            do j=1,n
                a(j,i) = inpt_a(j:j)
            end do
        end do
        do i=1,m
            read*, inpt_b
            do j=1,m
                b(j,i) = inpt_b(j:j)
            end do
        end do
    end block input_char

    do i=1,n-m+1
        do j=1,n-m+1
            if (fit(i,j))then
                print'(a)', 'Yes'
                stop
            end if
        end do
    end do

    print'(a)', 'No'

contains
    function fit(fx,fy) result(ret)
        integer(int32):: fx,fy
        integer(int32):: x,y
        logical:: ret   
        ret = .true.
        do x=1,m
            do y=1,m
                if (b(x,y) /= a(x+fx-1,y+fy-1)) then
                    ret = .false.
                    return
                end if
            end do
        end do
    end function
end program abc054b