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>
#define SORT(a) sort(a.begin(), a.end())
#define RSORT(a) sort(a.rbegin(), a.rend())
#define REP(i, n) for (int i = 0; i < n; i++)
#define RREP(i, n) for (int i = n - 1; 0 <= i; i--)
#define FOR(i, start, end) for (int i = start; i < end; i++)
#define RFOR(i, start, end) for (int i = start - 1; 0 <= i; i--)
#define ALL(a) a.begin(), a.end()
using ll = long long;
using namespace std;
const int INF32       = 1'050'000'000;
const long long INF64 = 4'000'000'000'000'000'000;
const int MOD7        = 1'000'000'007;
const int MOD9        = 1'000'000'009;
template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
void print() { std::cout << '\n'; }
template <class H, class... T> void print(H &&head, T &&... args) {
    std::cout << head;
    sizeof...(args) == 0 ? std::cout << "" : std::cout << ' ';
    print(std::forward<T>(args)...);
}
template <class T> void print(std::vector<T> &v) {
    for (int i = 0; i < v.size(); i++) {
        std::cout << v[i];
        i == v.size() - 1 ? std::cout << '\n' : std::cout << ' ';
    }
}
template <class T> void print(std::vector<std::vector<T>> &v) {
    for (int i = 0; i < v.size(); i++) {
        for (int j = 0; j < v[i].size(); j++) {
            std::cout << v[i][j];
            j == v[i].size() - 1 ? std::cout << '\n' : std::cout << ' ';
        }
    }
}

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

    int N, M;
    cin >> N >> M;
    vector<string> a(N), b(M);
    for (auto &&i : a) {
        cin >> i;
    }
    for (auto &&i : b) {
        cin >> i;
    }
    REP(i, N - M + 1) {
        REP(j, N - M + 1) {
            bool f = true;
            REP(y, M) {
                REP(x, M) {
                    if (a[y + i][x + j] != b[y][x]) {
                        f = false;
                        break;
                    }
                }
            }
            if (f) {
                print("Yes");
                return 0;
            }
        }
    }
    print("No");

    return 0;
}