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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*
import kotlin.math.acos
import kotlin.math.cos
import kotlin.math.round
import kotlin.math.sin

fun main(args: Array<String>) {
    val (n, m) = readIntList()
    val s = readString()
    val t = readString()

    if (s[0] != t[0]) {
        println(-1)
        return
    }
    if (n % m == 0 || m % n == 0) {
        println(-1)
        return
    }
    println(lcm(n.toLong(), m.toLong()))
}

// read
fun readString() = readLine()!!
fun readInt() = readLine()!!.toInt()
fun readLong() = readLine()!!.toLong()
fun readStringList() = readLine()!!.split(" ")
fun readIntList() = readLine()!!.split(" ").map(String::toInt)
fun readLongList() = readLine()!!.split(" ").map(String::toLong)

// 最大公約数
fun gcd(p: Long, q: Long): Long = if (p % q == 0L) q else gcd(q, p % q)

// 最小公倍数
fun lcm(p: Long, q: Long): Long = p / gcd(p, q) * q