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)

#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <tuple>
#include <numeric>
#include <string>
#include <bitset>
#include <functional>

using ld = long double;
using vld = std::vector<ld>;
using vvld = std::vector<vld>;
using vvvld = std::vector<vvld>;
using ll = long long;
using vll = std::vector<ll>;
using vvll = std::vector<vll>;
using vvvll = std::vector<vvll>;
using pll = std::pair<ll, ll>;
using vpll = std::vector<pll>;
using str = std::string;
using vstr = std::vector<str>;

template<typename T> bool chmin(T a, T &b) { return a < b ? b = a, true : false; }
template<typename T> bool chmax(T a, T &b) { return a > b ? b = a, true : false; }

constexpr ld PI = 3.141592653589793L;
constexpr ld EPS = std::numeric_limits<ld>::epsilon();

ll gcd(ll n, ll m) { return m ? gcd(m, n % m) : n; }
ll lcm(ll n, ll m) { return n / gcd(n, m) * m; }
bool is_prime(ll n) {
    if (n == 1) return false;
    for (ll i = 2; i * i <= n; i++)
        if (n % i == 0) return false;
    return true;
}

/*------------------------------------------------------------------------*/

void solve();

int main() {
    std::cin.tie(0);
    std::ios::sync_with_stdio(false);
    std::cout << std::setprecision(15);

    solve();
}

void solve() {
    ll n, m;
    std::cin >> n >> m;

    vstr a(n), b(m);
    for (ll i = 0; i < n; i++) std::cin >> a[i];
    for (ll i = 0; i < m; i++) std::cin >> b[i];

    for (ll i = 0; i < n - m + 1; i++) for (ll j = 0; j < n - m + 1; j++) {
        bool flg = true;
        for (ll k = 0; k < m; k++) for (ll l = 0; l < m; l++)
            if (a[i + k][j + l] != b[k][l]) { flg = false; break; }
        if (flg) { std::cout << "Yes" << '\n'; return; }
    }

    std::cout << "No" << '\n';
}