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: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
using System.IO;
using System.Linq;
using System.Diagnostics;
using System.Globalization;
using static System.Console;
using static System.Math;

namespace agc28_a
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] input = ReadLine().Split().Select(int.Parse).ToArray();
            string s = Console.ReadLine();
            string t = Console.ReadLine();

            int n = input[0];
            int m = input[1];

            //var list1 = new List<int>();
            //var list2 = new List<int>();

            //for (int i = 1; i*i <= n; i++)
            //{
            //    if(n % i == 0){
            //        list1.Add(i);
            //    }
            //}

            //for (int i = 1; i*i <= m; i++)
            //{
            //    if(m % i == 0){
            //        list2.Add(i);
            //    }
            //}

            int lcm = Lcm(n,m);

            int ln = lcm/n;
            int lm = lcm/m;
            Dictionary<int, char> dic = new Dictionary<int, char> ();

            for (int i = 0; i < n; i++){
                dic[ln*i+1] = s[i];
            }

            bool judge = true;
            for (int i = 0; i < m; i++){
                if(dic.ContainsKey(lm*i+1)){
                    if(dic[lm*i+1] != t[i]){
                        judge = false;
                    }
                }
            }

            if(judge){
                WriteLine(lcm);
            }else{
                WriteLine(lcm);
            }
        }

        static int Gcd(int a, int b) {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            while (b != 0) {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }

        public static int Lcm(int a, int b) {
            return a * b / Gcd(a, b);
        }
    }
}