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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <string.h>
#include <algorithm>

using namespace std;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)

int main(){
    int N,M;
    cin >> N >> M;
    char A[N][N],B[M][M];
    FOR(i,0,N){
        string s;
        cin >> s;
        strcpy(A[i], s.c_str());
    }
    FOR(i,0,M){
        string s;
        cin >> s;
        strcpy(B[i], s.c_str());
    }
    bool b;
    FOR(i,0,N-M+1){
        FOR(j,0,N-M+1){
            b=true;
            FOR(k,0,M){
                FOR(l,0,M){
                    if(A[i+k][j+l]!=B[k][l]){
                        b=false;
                        k=M-1;
                        break;
                    }
                }
            }
            if(b) {
                cout << "Yes" << endl;
                i=N-M;
                break;
            }
        }
    }
    if(!b) cout << "No" << endl;
    return 0;
}