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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS

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

/* define */
#define Yes(X) puts ( X ? "Yes" : "No")
#define YES(X) puts ( X ? "YES" : "NO")
#define FOR(i, a, b) for ( int i = a; i < b ; i++)
#define REP(i, n) for ( int i = 0; i < n ; i++)
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max3(a, b, c) ((max((a), (b)) > (c)) ? max((a), (b)) : (c))
#define min3(a, b, c) ((min((a), (b)) < (c)) ? min((a), (b)) : (c))

/* const */
const int MOD = (int)1e9 + 7;

/* alias */
typedef long long ll;
typedef unsigned  u;

/* function */
int cmp(const void *a, const void *b) {
	return *(int*)a - *(int*)b;
}
int cmp2(const void *a, const void *b) {
	return *(int*)b - *(int*)a;
}
int gcd(int a, int b) {
	if (b != 0)return gcd(b, a % b);
	return a;
}
ll Lgcd(ll a, ll b) {
	if (b != 0) return Lgcd(b, a % b); 
	return a;
}
int lcm(int a, int b) {
	return a / gcd(a, b) * b;
}
ll Llcm(ll a, ll b) {
	return a / Lgcd(a, b) * b;
}

/* main */

int main() {
	int n, m, c = 0;
	char a[50][100], b[50][100];
	scanf("%d%d", &n, &m);
	REP(i, n)scanf("%s", a);
	REP(i, m)scanf("%s", b);
	REP(i, n - m + 1) {
		REP(j, n - m + 1) {
			int f = 1;
			REP(k, m) {
				REP(l, m) {
					if (a[i + k][j + l] != b[k][l])f = 0;
				}
			}
			if (f)c = 1;
		}
	}
	Yes(c);
	return 0;
}