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)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn gcd(a: u64, b: u64) -> u64 {
    if b == 0 {
        a
    } else {
        gcd(b, a % b)
    }
}

fn lcm(a: u64, b: u64, g: u64) -> u64 {
    a / g * b
}

fn main() {
    input! {
        n: u64,
        m: u64,
        s: String,
        t: String,
    }

    let vs: Vec<char> = s.chars().collect();
    let vt: Vec<char> = t.chars().collect();
    let g = gcd(n, m);
    let l = lcm(n, m, g);

    // 理解しやすいが計算量の増える別解: LCM(N/L, M/L) の倍数の位置すべてを確認する
    // https://betrue12.hateblo.jp/entry/2018/10/14/121018
    // GCD が分母にくる理由を端折らずに解説してくださる良心
    // https://microkents.hatenablog.com/entry/2018/12/30/145128
    // 双方の文字列が進んでいく上で進み幅 L/N と L/M の公倍数となる位置がぶつかる
    // 例えば s については LCM(L/N, L/M) 動くと公倍数となり、文字一つで動く量は L/N
    // LCM(L/N, L/M) / (L/N) = L^2/NM * N/L = L/M
    for i in 0..g {
        // if vs[(i * (n / g)) as usize] != vt[(i * (m / g)) as usize] {
        if vs[(i * (l / m)) as usize] != vt[(i * (l / n)) as usize] {
            println!("-1");
            return;
        }
    }

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