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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;

class Amondai
{
    static void Main()
    {
        var num = Console.ReadLine().Split(' ').Select(x => Int64.Parse(x)).ToArray();
        var max = num[0] < num[1] ? num[1] : num[0];
        var min = max == num[0] ? num[1] : num[0];
        var ans = -1L;
        var line1 = Console.ReadLine().ToCharArray();
        var line2 = Console.ReadLine().ToCharArray();
        if (line1[0] != line2[0])
        {
            Console.WriteLine(ans);
            return;
        }
        ans = GetS(max, min);
        if (max * min != ans)
        {
            var kankaku1 = ans / num[0];
            var kankaku2 = ans / num[1];
            var kankaku = GetS(kankaku1, kankaku2);
            for (long i = kankaku; i < ans; i += kankaku)
            {
                if (line1[i / kankaku1] != line2[i / kankaku2] || (line1.Count() < i / kankaku1) || (line2.Count() < i / kankaku2))
                {
                    ans = -1;
                    break;
                }
            }
        }
        Console.WriteLine(ans);
    }

    static long GetS(long max, long min)
    {
        var r = 0L;
        var a = max;
        var b = min;
        while ((r = max % min) != 0)
        {
            max = min;
            min = r;
        }
        return a * b / min;
    }
}