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 3.8.0) (judge verdict: Accepted)

// ABC054 B - Template Matching
#include <stdio.h>
#include <string.h>

int main(void) {
    int N, M;
    scanf("%d %d", &N, &M);
    char A[50][51];
    char B[50][51];
    char joinA[2501] = {'\0'};
    char joinB[2501] = {'\0'};
    int i;
    for (i = 0; i < N; i++) {
        scanf("%s", A[i]);
    }
    for (i = 0; i < M; i++) {
        scanf("%s", B[i]);
    }

    for (i = 0; i < M; i++) {
        strcat(joinB, B[i]);
    }
    int x, y;
    for (y = 0; y < (N - M + 1); y++) {
        for (x = 0; x < (N - M + 1); x++) {
            if (A[y][x] == joinB[0]) {
                // joinAを作る
                for (i = 0; i < M; i++) {
                    strncat(joinA, &A[y + i][x], M);
                }
                // printf("joinA:%s\n",joinA);
                // joinA=joinBならYes
                if (strcmp(joinA, joinB) == 0) {
                    printf("Yes\n");
                    return 0;
                }
                joinA[0] = '\0';
            }
        }
    }

    printf("No\n");
    return 0;

}