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: Wrong Answer)

#include<bits/stdc++.h>

#define fi first
#define se second
#define FO(x, n) for (int x = 0; x < n; ++x)
#define FOR(x, a, b) for (int x = a; x < b; ++x)
#define RFO(x, n) for (int x = n - 1; x >= 0; --x)
#define RFOR(x, a, b) for (int x = b - 1; x >= a; --x)
#define FOR_ITER(x, a) for(auto x = a.begin(); x != a.end(); ++x)



using namespace std;
typedef unsigned char byte;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<string, string> pss;
typedef pair<ll, ll> pll;
inline bool feq(const double& a, const double& b) { return fabs(a - b) < 1e-10; }

using namespace std;

int n, m;
vector<string> a;
vector<string> b;

bool cmp(int idx){
    int cIdx = -1;
    int cId;
    FOR(i, idx, (idx+m)){
        cId = a.at(i).find(b.at(i - idx));
        if(cId == string::npos){
            return false;
        }

        if (cIdx == -1){
            cIdx = cId;
        } else {
            if(cIdx != cId) {
                return false;
            }
        }
    }
    return true;
}

int main(){
//    int n, l;
//    cin >> n >> k;
//    vector<bool> dislike(10, false);
//
//    FOR(i, 0, k) {
//        int t;
//        cin >> t;
//        dislike[t] = true;
//    }



cin >>n>>m;
a.resize(n);
b.resize(m);

FO(i, n){
    cin >> a.at(i);
}
FO(i, m){
    cin >> b.at(i);
}

int cur = 0;

FO(i, (n - m + 1)) {
    if(cmp(i)) {
        cout<<"Yes"<< endl;
        return 0;
    }
}

cout<<"No"<<endl;

    return 0;
}