Problem:
Score : 300 points

Problem Statement

You are given a string S of length N and another string T of length M.
These strings consist of lowercase English letters.

A string X is called a good string when the following conditions are all met:

Let L be the length of X. L is divisible by both N and M.

Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.

Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.

Determine if there exists a good string. If it exists, find the length of the shortest such string.

Constraints

1 \leq N,M \leq 10^5

S and T consist of lowercase English letters.

|S|=N

|T|=M

Input

Input is given from Standard Input in the following format:

N M
S
T

Output

If a good string does not exist, print -1; if it exists, print the length of the shortest such string.

Sample Input 1

3 2
acp
ae

Sample Output 1

6

For example, the string accept is a good string.
There is no good string shorter than this, so the answer is 6.

Sample Input 2

6 3
abcdef
abc

Sample Output 2

-1

Sample Input 3

15 9
dnsusrayukuaiia
dujrunuma

Sample Output 3

45

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn read() -> String {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).ok();
    buf
}

fn gcd(a: usize, b: usize) -> usize {
    let mut x = a;
    let mut y = b;
    while y != 0 {
        let tmp = y;
        y = x % y;
        x = tmp;
    }
    x
}

fn main() {
    let (n, m) = {
        let v: Vec<usize> = read()
            .split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect();
        (v[0], v[1])
    };
    let s: Vec<char> = read().trim().chars().collect();
    let t: Vec<char> = read().trim().chars().collect();

    let gcd = gcd(n, m);
    if (0..gcd).all(|i| s[n / gcd * i] == t[m / gcd * i]) {
        println!("{}", n * m / gcd);
    } else {
        println!("-1");
    }
}