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;//リストの使用
using System.Collections.Generic;
class Program
{
	static void Main()
	{
		string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
		long n = long.Parse(input[0]);
		long m = long.Parse(input[1]);
		string s = Console.ReadLine();
    string t = Console.ReadLine();
    bool answer = true;
    long memo = 0;

    long gcdNM = Gcd(n,m);
    long lcmNM = n * m / gcdNM;//最小公倍数

    if(s[0] != t[0]) answer = false;
    if(n < m)
    {
      long swapMemo = n;
      n = m;
      m = swapMemo;
      string stringMemo = s;
      s = t;
      t = stringMemo;
    }

    if(n % m == 0)
    {
      memo = n/m;
      for(long i = 1; i < m; i++)
      {
        if(s[(int)(memo*i)] != t[(int)i]) answer = false;
      }
    }

    else if(gcdNM > 1)
    {
      long a = lcmNM/n;
      long b = lcmNM/m;
      long gcdAB = Gcd(a,b);
      long lcmAB = a*b/gcdAB;
      for(long i = 1; i < lcmNM/lcmAB; i++)
      {
        long c = lcmAB / a;
        long d = lcmAB / b;
        if(s[(int)(c*i)] != t[(int)(d*i)]) answer = false;
      }
    }
    
		Console.WriteLine((answer) ? lcmNM : -1);
	}

  static long Gcd(long a, long b) 
    {//引数a,bの最大公約数を返す
         if (a < b)
             return Gcd(b, a);//入れ替え
         while (b != 0) 
         {
             long changeTo = a % b;
             a = b;
             b = changeTo;
         }
         return a;
     }
  
}