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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define MOD = 1000000007;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define ALL(a) begin(a),end(a)
#define COUNTOF(a) (sizeof(a)/sizeof(a[0]))
template<class T> istream& operator >> (istream& s, vector<T>& v) {
  for (auto& x: v) { s >> x; } return s;}
template<class T> inline bool chmin(T& a, T b){return a>b?a=b,true:false;}
template<class T> inline bool chmax(T& a, T b){return a<b?a=b,true:false;}
#define SP << " " <<
#define FMT(a) #a<<":"<<a 
#define FMT2(a,b) #a<<":"<<a<<", "<<#b<<":"<<b 
#define FMT3(a,b,c) #a<<":"<<a<<", "<<#b<<":"<<b<<", "<<#c<<":"<<c 
#define FMT4(a,b,c,d) #a<<":"<<a<<", "<<#b<<":"<<b<<", "<<#c<<":"<<c<<", "<<#d<<":"<<d 
#define ANS(a) cout<<a<<endl
void yes(bool f){cout<<(f?"yes":"no")<<endl;}
void Yes(bool f){cout<<(f?"Yes":"No")<<endl;}
void YES(bool f){cout<<(f?"YES":"NO")<<endl;}

int main()
{
    cin.tie(0); ios::sync_with_stdio(false);
    //cout<<std::fixed<<std::setprecision(10);

    int N, M; cin >> N >> M;
    vector<string> A(N); cin >> A;
    vector<string> B(M); cin >> B;

    // cout << FMT2(N, M) << endl;
    for (int i = 0; i < N - M + 1; i++) {
        for (int j = 0; j < N - M + 1; j++) {
            // cout << FMT2(i, j) << endl;
            bool f = true;
            for (int k = 0; k < M; k++) {
                // cout SP FMT3(k, B[i + k], A[i + k].substr(j, M)) << endl;
                if (B[i + k] != A[i + k].substr(j, M)) {
                    f = false;
                    break;
                }
            }
            if (f) {
                Yes(true);
                return 0;
            }
        }
    }
    Yes(false);
}