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 (1.0.0) (judge verdict: Time Limit Exceeded)

fun main(args : Array<String>) {
    val (n, m) = readLine()!!.split(" ").map { it.toInt() }
    val s = readLine()!!
    val t = readLine()!!

    val lcm = lcm(n.toLong(), m.toLong())
    val stepN = (lcm/n).toInt()
    val stepM = (lcm/m).toInt()

    var nIdx = 0L
    var mIdx = 0L

    var nCnt = 0
    var mCnt = 0

    for (i in 0L until lcm) {
        if (i == nIdx && i == mIdx) {
            if (s[nCnt] != t[mCnt]) {
                println("-1")
                return
            }

            nCnt++
            mCnt++

            nIdx = if(nCnt < n) nIdx + stepN else -1
            mIdx = if(mCnt < m) mIdx + stepM else -1
        } else if (i == nIdx) {
            nCnt++
            nIdx = if(nCnt < n) nIdx + stepN else -1
        } else if (i == mIdx) {
            mCnt++
            mIdx = if(mCnt < m) mIdx + stepM else -1
        }
    }

    println(lcm)
}

fun gcd(a : Long, b : Long) : Long {
    return if (a % b == 0L) b else gcd(b, a % b)
}

fun lcm(a : Long, b : Long) : Long {
    return (a * b) / gcd(a, b)
}