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 C (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main(void) {
	int i, j, k ,l;
	int N, M;
	int result;
	int kl_break;
	char **a, **b;
	char buf[128], *ch;

	fgets(buf, sizeof(buf), stdin);
	ch = strtok(buf, " ");
	N = atoi(ch);
	ch = strtok(NULL, " ");
	M = atoi(ch);

	a = (char **)calloc(N, sizeof(char *));
	for(i = 0; i < N; i += 1) {
		a[i] = (char *)calloc(64, sizeof(char *));
	}
	b = (char **)calloc(N, sizeof(char *));
	for(i = 0; i < M; i += 1) {
		b[i] = (char *)calloc(64, sizeof(char *));
	}

	for(i = 0; i < N; i += 1) {
		fgets(a[i], sizeof(buf), stdin);
	}
	for(i = 0; i < M; i += 1) {
		fgets(b[i], sizeof(buf), stdin);
	}

	result = 0;
	for(i = 0; i < N - M + 1; i += 1) {
		for(j = 0; j < N - M + 1; j += 1) {
			kl_break = 0;
			//bの左上はaのi,jにある
			for(k = 0; k < M; k += 1) {//bと(aの一部)の一致を調べる
				for(l = 0; l < M; l += 1) {//bと(aの一部)の一致を調べる
					if(a[i + l][j + k] != b[l][k]) {
						kl_break = 1;
						break;//if 1 pixel different exit kl loop
					}
				}
				if(kl_break == 1) {//if 1 pixel different exit kl loop
					break;
				}
			}
			if(kl_break == 0) {//if all pixels same in kl loop exit ij loop
				result = 1;
				break;
			}
		}
		if(result == 1) {
			break;
		}
	}

	if(result == 1) {
		printf("Yes\n");
	} else {
		printf("No\n");
	}

	for(i = 0; i < N; i += 1) {
		free(a[i]);
	}
	free(a);
	for(i = 0; i < M; i += 1) {
		free(b[i]);
	}
	free(b);

	return 0;
}