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>
#include<algorithm>
#include<math.h>
#include<string>
#include<iostream>
#include<vector>
#include<cstdlib>
#include<stdio.h>
#include<cmath>
#include<limits>
#include<iomanip>
#include<cstdlib> 
#include<ctime>
#include<random>


using namespace std;

template <class T> using V = vector<T>;

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;
}

const long long INF = 1LL << 60;

using ll = long long;
using db = long double;
using st = string;
using ch = char;
using vll = V<ll>;
using vpll =V<pair<ll,ll>>;
using vst = V<st>;
using vdb = V<db>;
using vch = V<ch>;
using graph = V<V<int>>;
using pq = priority_queue<ll>;



#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define bgn begin()
#define en end()
#define SORT(a) sort((a).bgn,(a).en)
#define REV(a) reverse((a).bgn,(a).en)
#define fi first
#define se second
#define sz size()
#define gcd(a,b) __gcd(a,b)
#define pb(a) push_back(a);
#define ALL(a)  (a).begin(),(a).end()




ll Sum(ll n) {
ll m=0;
  while(n){
    m+=n%10;
    n/=10;
  }
  return m;
}



const int MAX = 510000;
const int MOD = 1000000007;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
void Comuse() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}


#define comuse Comuse()

ll combi(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll perm(int n,int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] % MOD) % MOD;
}

ll modpow(ll a,ll n,ll mod){
  ll ans=1;
  while(n>0){
    if(n&1){
      ans=ans*a%mod;
    }

    a=a*a%mod;
    
    n>>=1;
  }

  return ans;
}

ll modinv(ll a, ll mod) {
    return modpow(a, mod - 2, mod);
}

ll modcombi(int n,int k,int mod){
  ll ans=1;
  for(ll i=n;i>n-k;i--){
    ans*=i;
    ans%=mod;
  }
  for(ll i=1;i<=k;i++){
    ans*=modinv(i,mod);
    ans%=mod;
  }

  return ans;
}

ll lcm(ll a,ll b){
  ll n;
  n=a/gcd(a,b)*b;
  return n;
}


vll div(ll n){
  vll ret;
  for(ll i=1;i*i<=n;i++){
    if(n%i==0){
      ret.push_back(i);
      if(i*i!=n){
        ret.push_back(n/i);
      }
    }
  }
  SORT(ret);
  return (ret);
}

vector<bool> isprime(MAX+100,true);

void primeuse(){
  isprime[0]=false;
  isprime[1]=false;
  for(int i=2;i<MAX+50;i++){
    int up=sqrt(i)+1;
    for(int j=2;j<up;j++){
      if(i%j==0){
        isprime[i]=false;
      }
    }
  }
}



void Solve();


const int MAX_N = 131072;
//segment tree 
int NN;
int seg[MAX_N*2-1];
void seguse(){
  for(int i=0;i<2*NN-1;i++){
    seg[i]=INT_MAX;
  }
}


signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<setprecision(20)<<fixed;
  Solve();
}



/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/
//segtreeの葉の先頭の添え字はN-1
void Solve(){
  //ll N=262144;
  //vll segtree(2*N-1);
  ll n,m;
    cin>>n>>m;
  vector<string> A(n);
  FOR(i,0,n){
    cin>>A[i];
  }
  vector<string> B(m);
  FOR(i,0,m){
    cin>>B[i];
  }

  bool ans=false;
  FOR(i,0,n-m+1){
    FOR(j,0,n-m+1){
      //左上の全探索(i,j)
      bool how=true;
      FOR(k,j,j+m){
        string now=A[k].substr(i,m);
        if(now!=B[k-j]){
          how=false;
        }
      }
      if(how){
        ans=true;
      }
    }
  }

  if(ans){
    cout<<"Yes\n";
  }
  else{
    cout<<"No\n";
  }
}