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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import "fmt"

func main() {
	// Code for B - Template Matching
	var N, M int
	fmt.Scanf("%d %d", &N, &M)

	A := make([]string, N)
	B := make([]string, M)

	for i := 0; i < N; i++ {
		fmt.Scanf("%s", &A[i])
	}

	for i := 0; i < M; i++ {
		fmt.Scanf("%s", &B[i])
	}

	var flag1, flag2 int
	for i := 0; i < (N - M + 1); i++ {
		for j := 0; j < (N - M + 1); j++ {
			flag1 = 0
			for k := 1; k <= M; k++ {
				if A[i][j:j+k] != B[0][0:k] {
					flag1 = 1
				}
			}
			if flag1 == 0 {
				flag2 = 1
				break
			}
		}
	}

	if flag2 == 0 {
		fmt.Println("No")
	} else {
		fmt.Println("Yes")
	}
}