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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REP(i, n) for(ll i = 0; i < (ll)(n); i++)
#define FOR(i, a, b) for(ll i = a; i <= (ll)(b); i++)
#define INF 1e9
#define endl '\n'

int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);

    int N, M;
    cin >> N >> M;

    char AA[N][N];
    char BB[M][M];

    REP(y, N) {
        REP(x, N) { cin >> AA[y][x]; }
    }

    REP(y, M) {
        REP(x, M) { cin >> BB[y][x]; }
    }

    REP(x0, N - M + 1) {
        REP(y0, N - M + 1) {
            bool tmp = true;
            REP(x, M) {
                REP(y, M) {
                    if(AA[y0 + y][x0 + x] != BB[y][x]) {
                        tmp = false;
                    }
                }
            }
            if(tmp) {
                cout << "Yes" << endl;
                return 0;
            }
        }
    }

    cout << "No" << endl;
}