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;

#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define rept(i,n) for(int (i)=0;(i)<=(int)(n);(i)++)
#define reps(i,s,n) for(int (i)=(s);(i)<(int)(n);(i)++)
#define repst(i,s,n) for(int (i)=(s);(i)<=(int)(n);(i)++)
#define repr(i,n) for(int (i)=(n);(i)>=0;(i)--)
#define each(itr,v) for(auto (itr):(v))
#define all(c) (c).begin(),(c).end()
#define pb push_back
#define mp(x,y) make_pair((x),(y))
#define fi first
#define se second
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define ln "\n"
#define show(x) cout << #x << " = " << x ln
#define dbg(x) cout<<#x"="<<x ln

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vector<int> > mat;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const int inf = (int)1e9;
const ll linf = (ll)1e18;
const int mod = (int)(1e9+7);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};

struct oreno_initializer {
	oreno_initializer() {
		cin.tie(0);
		ios::sync_with_stdio(0);
	}
} oreno_initializer;

// ━━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…
// .｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋
// ・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・

int n, m;
string s[50], t[50];

int main() {
	cin >> n >> m;
	rep(i,n) cin >> s[i];
	rep(i,m) cin >> t[i];
	
	rept(ii,n-m) {
		rept(jj,n-m) {
			rep(i,m) {
				rep(j,m) {
					if (s[ii+i][jj+j]!=t[i][j]) goto br;
				}
			}
			cout << "Yes" << ln;
			return 0;
			br:;
		}
	}
	cout << "No" << ln;
}