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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
	input!(n: usize);
	input!(m: usize);
	input!(a: [String; n]);
	input!(b: [String; m]);

	let mut an: [u64; 50] = [0; 50];
	for i in 0..n {
		for j in 0..n {
			an[i] <<= 1;
			if a[i].chars().nth(j).unwrap() == '#' {
				an[i] += 1;
			}
		}
	}

	let mut bn: [u64; 50] = [0; 50];
	for i in 0..m {
		for j in 0..m {
			bn[i] <<= 1;
			if b[i].chars().nth(j).unwrap() == '#' {
				bn[i] += 1;
			}
		}
	}

	for y in 0..n-m+1 {
		for x in 0..n-m+1 {
			let mut ng_flag = 0;
			for u in 0..m {
				let at = an[y+u] << (64-n+x) >> (64-m);
				if at != bn[u] {
					ng_flag = 1;
					break;
				}
			}
			if ng_flag == 0 {
				println!("Yes");
				return;
			}
		}
	}
	println!("No");
}