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 9.2.1) (judge verdict: Accepted)

// https://atcoder.jp/contests/abc054/tasks/abc054_b
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define REP(i, n) FOR(i, 0, n)
#define FOR(i, s, n) for (int i = (s), i##_len = (n); i < i##_len; ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
#define CEIL(a, b) ((a - 1) / b + 1)

void solve() {
    int N, M;
    cin >> N >> M;
    vector<string> A(N), B(M);
    for (auto &x : A) {
        cin >> x;
    }
    for (auto &x : B) {
        cin >> x;
    }
    REP(h, N - M + 1) REP(w, N - M + 1) {
        bool ng = false;
        REP(i, M) REP(j, M) {
            if (A[h + i][w + j] != B[i][j]) {
                ng = true;
                break;
            }
        }
        if (!ng) {
            cout << "Yes\n";
            return;
        }
    }
    cout << "No\n";
}

signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}