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::collections::BTreeMap;
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
fn main(){
    let (n,m) = {
        let t = read_vec::<usize>();
        (t[0],t[1])
    };
    let s:Vec<char> = read::<String>().chars().collect();
    let t:Vec<char>  = read::<String>().chars().collect();
    let lcmv = lcm(n,m);
    let mut mm = BTreeMap::new();
    for (i,c) in s.into_iter().enumerate(){
        mm.insert(lcmv/n*i,c);
    }
    let mut check_flag = true;
    for (i,c) in t.into_iter().enumerate(){
        if let Some(x) = mm.get(&(lcmv/m * i)){
            if c != *x {
                check_flag = false;
            }
        }
    }
    let ans = if check_flag {lcmv as i64} else { -1 };
    println!("{}",ans);
}

fn gcd(x:usize,y:usize)-> usize{
    if x % y == 0{
        y
    } else{
        gcd(y,x%y)
    }
}
fn lcm(x:usize,y:usize)->usize{
    x*y/gcd(x,y)
}