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::cmp::{max, min};
use std::collections::HashSet;
use std::process::exit;
const MOD: usize = 1000000007;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    // var... 変数の識別子, $t...型を一つよむ
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        //ここで繰り返し
        input_inner!{$iter $($r)*}
    };
}

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<_>>()
    };

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

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    // 配列の最後のNestではここで型が指定されてparseされる
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}
// =========

fn gcd(a: usize, b: usize) -> usize {
    if b == 0 {
        return a;
    } else if a < b {
        return gcd(a, b % a);
    } else {
        return gcd(b, a % b);
    }
}

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

    let lcm = n * m / gcd(n, m);
    let mut nfac = HashSet::new();
    let mut mfac = HashSet::new();
    for i in 0..n {
        nfac.insert(lcm / n * i);
    }
    for i in 0..m {
        mfac.insert(lcm / m * i);
    }
    for i in nfac.intersection(&mfac) {
        if s[i * n / lcm] != t[i * m / lcm] {
            println!("{:?}", -1);
            exit(0);
        }
    }
    println!("{:?}", lcm);
}