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)

// Template from https://techracho.bpsinc.jp/jhonda/2019_08_05/78537

// input macro for competive-programming from https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };

    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, bytes) => {
        read_value!($iter, String).into_bytes()
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    use std::io::Read;
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();

    println!("{}", solve(&s));
}

macro_rules! test {
    ($($input:expr => $output:expr),* $(,)*) => {
        #[test]
        fn solve_test() {
            $(
                assert_eq!(solve($input), $output);
            )*
        }
    };
}

test! {
r#"3 2
#.#
.#.
#.#
#.
.#"# => r#"Yes"#,

r#"4 1
....
....
....
....
#"# => r#"No"#,

r#"3 2
###
.#.
#.#
#.
.#"# => r#"Yes"#,

r#"3 2
###
.##
#.#
#.
.#"# => r#"No"#,

}

fn solve(src: &str) -> String {
    input! {
        source = src,
        n: usize, m: usize,
        a: [chars; n],
        b: [chars; m]
    }
    println!("{:?}", a);
    println!("{:?}", b);
    let mut ans = "No";

    'main: for ax in 0..n {
        'sub: for ay in 0..n {
            if a[ax][ay] != b[0][0] {
                continue;
            }
            // println!("--------------------");


            let mut cnt = 0;
            for bx in 0..m {
                for by in 0..m {
                    // print!("a[{}][{}]={}, b[{}][{}]={}:", ax + bx, ay + by, a[ax + bx][ay + by], bx, by, b[bx][by]);
                    if ax + bx >= n || ay + by >= n || a[ax + bx][ay + by] != b[bx][by] {
                        // println!(" ng");
                        continue 'sub;
                    }

                    cnt += 1;
                    // println!(" ok");
                }
            }

            if cnt == m * m {
                ans = "Yes";
                break 'main;
            }
        }
    }

    ans.to_string()
}