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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.ArrayDeque

fun main(args: Array<String>){
    val (N, M) = readLine()!!.split(" ").map { it.toInt() }
    val A = Array(N){ readLine()!! }
    val B = Array(M){ readLine()!! }


    val queue = ArrayDeque<Triple<Int, Int, Int>>()

    for(i in 0..N-M){
        if(A[i].contains(B[0])){
            queue.add(Triple(i, A[i].indexOf(B[0]), 0))
        }
    }
    while (queue.isNotEmpty()){
        val next = queue.poll()
        if(next.third + 1 == M){
            println("Yes")
            return
        }else{
            if(A[next.first + 1].contains(B[next.third + 1]) && A[next.first + 1].indexOf(B[next.third + 1]) == next.second){
                queue.add(Triple(next.first + 1, next.second, next.third + 1))
            }
        }
    }
    println("No")
}