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)

use std::io::Read;
use std::collections::HashMap;

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

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let m: usize = iterator.next().unwrap().parse().unwrap();
    let s: Vec<char> = iterator.next().unwrap().chars().collect();
    let t: Vec<char> = iterator.next().unwrap().chars().collect();

    let lcm = lcm(n, m);
    // let mut x: Vec<char> = vec!['*'; lcm];
    let mut x: HashMap<usize, char> = HashMap::new();
    for i in 0..n {
        // x[lcm / n * i] = s[i];
        x.insert(lcm / n * i, s[i]);
    }
    for j in 0..m {
        // if x[lcm / m * j] != '*' && x[lcm / m * j] != t[j] {
        //     return "-1".to_string();
        // }
        if x.contains_key(&(lcm / m * j)) && *x.get(&(lcm / m * j)).unwrap() != t[j] {
            return "-1".to_string();
        }
    }
    // for i in 0..n {
    //     for j in 0..m {
    //         if lcm / n * i != lcm / m * j {
    //             continue;
    //         }
    //         if s[i] != t[j] {
    //             return "-1".to_string();
    //         }
    //     }
    // }

    return lcm.to_string();
}

fn lcm(m: usize, n: usize) -> usize {
    return m / gcd(m, n) * n;
}

fn gcd(mut m: usize, mut n: usize) -> usize {
    while n != 0 {
        let r = m % n;
        m = n;
        n = r;
    }
    m
}