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: Accepted)

import java.util.*

val sc = Scanner(System.`in`)
val A = mutableListOf<String>()
val B = mutableListOf<String>()
var N = 0
var M = 0

fun main(args: Array<String>) {
    N = sc.nextInt()
    M = sc.nextInt()

    for (i in 0 until N) {
        A.add(sc.next())
    }
    for (i in 0 until M) {
        B.add(sc.next())
    }
    for (i in 0 until N) {
        for (j in 0 until N) {
            if (judge(i, j)) {
                println("Yes")
                return
            }
        }
    }
    println("No")
}

fun judge(Ni: Int, Nj: Int): Boolean {
    for (i in 0 until M) {
        for (j in 0 until M) {
            when {
                Ni + i >= N || Nj + j >= N || B[i][j] != A[Ni + i][Nj + j]
                -> return false
            }
        }
    }
    return true
}