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: Runtime Error)

fn gcd<T>(a: T, b: T) -> T
where
    T: Copy + std::cmp::PartialEq + std::ops::Rem<Output = T> + std::convert::From<u8>,
{
    if b == T::from(0u8) {
        a
    } else {
        gcd(b, a % b)
    }
}

fn lcm<T>(a: T, b: T) -> T
where
    T: Copy
        + std::cmp::PartialEq
        + std::ops::Rem<Output = T>
        + std::ops::Mul<Output = T>
        + std::ops::Div<Output = T>
        + std::convert::From<u8>,
{
    a / gcd(a, b) * b
}

fn main() {
    proconio::input! {
        n: usize,
        m: usize,
        s: proconio::marker::Chars,
        t: proconio::marker::Chars,
    }
    let l = lcm(n, m);
    let mut x = vec!['.'; l];
    for i in 0..n {
        let idx = i * (l / n);
        x[idx] = s[i];
    }
    for i in 0..m {
        let idx = i * (l / m);
        if x[idx] == '.' || x[idx] == t[i] {
            x[idx] = t[i];
        } else {
            println!("-1");
            return;
        }
    }
    println!("{}", l);
}