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 (1.42.0) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

//https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

//最大公約数を求める
fn gcd(x: usize, y: usize) -> usize {
    if x == 0 {
        return y;
    } else {
        return gcd(y % x, x);
    }
}

fn main() {
    let n: usize = read();
    let m: usize = read();
    let s: Vec<char> = read::<String>().chars().collect();
    let t: Vec<char> = read::<String>().chars().collect();

    //最小公倍数を求める
    let l = n * (m / gcd(n, m));

    //文字列Xの候補に必要な要素のインデックスを得る
    let mut x_s: Vec<usize> = [1; 1].to_vec();
    let mut x_t: Vec<usize> = [1; 1].to_vec();
    for i in 1..n {
        x_s.push(x_s[i - 1] + (l / n));
    }
    for i in 1..m {
        x_t.push(x_t[i - 1] + (l / m));
    }

    //println!("x_s{:?}", x_s);
    //println!("x_t{:?}", x_t);

    let mut s_i = 0;
    let mut t_i = 0;
    let mut chk_s = vec![];
    let mut chk_t = vec![];

    //x_sとx_tで同じところを参照する箇所を抽出する
    while s_i != x_s.len() && t_i != x_t.len() {
        if x_s[s_i] == x_t[t_i] {
            chk_s.push(s_i);
            chk_t.push(t_i);
            s_i += 1;
            t_i += 1;
        } else {
            if x_s[s_i] < x_t[t_i] {
                s_i += 1;
            } else {
                t_i += 1;
            }
        }
    }

    //println!("chk_s{:?}", chk_s);
    //println!("chk_t{:?}", chk_t);

    let mut ans_flag = true;
    //同じところを参照して、一致しているか判定する
    for (x, y) in chk_s.into_iter().zip(chk_t.into_iter()) {
        if s[x] != t[y] {
            ans_flag = false;
        }
    }

    if ans_flag {
        println!("{}", l);
    } else {
        println!("-1");
    }
}