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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define rep(i, n) for (i = 0; i < n; i++)
#define rrep(i, n) for (i = n; i >= 0; i--)
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)

int eq(char a[50][50], int x, int y, char b[50][50], int s) {
    int i, j;
    rep(i, s) rep(j, s) {
        if (a[x + i][y + j] != b[i][j]) return 0;
    }
    return 1;
}

int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    getchar();

    char a[50][50], b[50][50];
    int i, j;
    rep(i, n) {
        rep(j, n) a[i][j] = getchar();
        getchar();
    }
    rep(i, m) {
        rep(j, m) b[i][j] = getchar();
        getchar();
    }

    rep(i, n - m + 1) rep(j, n - m + 1) {
        if (eq(a, i, j, b, m)) {
            puts("Yes");
            return 0;
        }
    }
    puts("No");
    return 0;
}