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 Scala (2.11.7) (judge verdict: Accepted)

import scala.io.StdIn

object Main extends App {

  val Array(n, m) = StdIn.readLine().split(" ").map(_.toInt)

  val a = (0 until n).map { _ =>
    StdIn.readLine()
  }

  val b = (0 until m).map { _ =>
    StdIn.readLine()
  }

  var im = 0
  var jm = 0
  var judge = false
  for (x <- 0 until n if (x + m) <= n && !judge){ 
    for (y <- 0 until n if (y + m) <= n && !judge) {
      judge = true
      for (z <- 0 until m if (z + m) <= n && judge) {
        //println(a(x + z).substring(y, y + m))
        //println(b(z))
        judge = a(x + z).substring(y, y + m) == b(z)
      }
    }
  }

  println(if (judge) "Yes" else "No")
}