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)

/** author: __nishad__ **/
//#pragma GCC optimize(2)
//#pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
using namespace std;


typedef long long ll ;
typedef vector<ll> vll ;

#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define all(v) (v).begin(),(v).end()
#define precision(k) cout<<setprecision(k)<<fixed;
#define loop(i,a,b) for(ll i=(a);i<(b);++i)
#define minv(v) *min_element(all(v))
#define maxv(v) *max_element(all(v))
#define getpos(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define unq(v) sort(all(v)),(v).erase(unique((v).begin(),(v).end()),(v).end())
#define pb push_back
#define mp make_pair
#define pi 3.14159265358979323
#define endl "\n"


char a[50][50], b[2500];
int n, m;


int solve(int i, int j)
{
    int count = 0, x=0;
    loop(o, i, i+m){
        loop(p, j, j+m){
            if(b[x] == a[o][p]){
                count++;
                x++;
            }

        }
    }
    return count;

}

int main()
{
    IOS;
    int count, x;

    cin>>n>>m;
    char s;

    loop(i, 0, n){
        loop(j, 0, n){
            cin>>a[i][j];
        }
    }

    loop(i, 0, m*m){
        cin>>b[i];
    }


    loop(i, 0, n){
        loop(j, 0, n){
            int res = 0;
            if(b[0] == a[i][j]){
                count = 0, x=0;
                loop(o, i, i+m){
                    loop(p, j, j+m){
                        if(b[x] == a[o][p]){
                            count++;
                            x++;
                        }

                    }
                }
            }
            if(count == m*m){
                cout<<"Yes";
                return 0;
            }
        }
    }

    cout<<"No";


}