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: Wrong Answer)

// lcmとか__builtin_popcountとかはg++ -std=c++17 default.cppみたいなかんじで
// g++ hoge.cpp -std=c++17 -I . でコンパイルできる
// -fsanitize=undefinedでオーバーフロー検出
#include <bits/stdc++.h>
//#include <atcoder/all>
#define mod 1000000007
#define INF LLONG_MAX
#define ll int
#define ln cout<<endl
#define Yes cout<<"Yes"<<endl
#define NO cout<<"NO"<<endl
#define YES cout<<"YES"<<endl
#define No cout<<"No"<<endl
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
using namespace std;
//using namespace atcoder;
typedef pair<ll,ll> P;
ll dx[4]={1,0,-1,0};
ll dy[4]={0,1,0,-1};


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

    ll a,b,c,d,m,n,k,x,y,maxi=0,f=0,sum=0;
    string str2,str1;
    int hoge=0;
    cin>>n>>m;
    string s1[n];
    string s2[m];
    rep(i,n) cin>>s1[i];
    rep(i,m) cin>>s2[i];
    for(ll i=0;i<n-m;i++){
        for(ll j=0;j<n-m;j++){
            f=0;
            for(ll k=0;k<m;k++){
                for(ll l=0;l<m;l++){
                    if(s1[i+k][j+l]!=s2[k][l])f=1;
                }
            }
            if(!f){
                Yes;
                return 0;
            }
        }
    }
No;


    return 0;
}