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 (1.15.1) (judge verdict: Wrong Answer)

fn main() {
    let n: usize = read();
    let m: usize = read();
    let a: Vec<Vec<char>> = (0..n).map(|_| read::<String>().chars().collect()).collect();
    let b: Vec<Vec<char>> = (0..m).map(|_| read::<String>().chars().collect()).collect();

    for i in 0..n - m + 1 {
        'outer: for j in 0..n - m + 1 {
            for k in 0..m {
                for l in 0..m {
                    if a[i + k][j + l] != b[k][l] {
                        break 'outer;
                    }
                }
            }
            println!("Yes");
            return;
        }
    }
    println!("No");
}


#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T {
    use std::io::prelude::*;

    let stdin = std::io::stdin();
    let s: String = stdin
        .lock()
        .bytes()
        .map(|b| b.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    s.parse().ok().unwrap()
}