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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mmax = 1e5 + 5;
char str1[mmax], str2[mmax];
int gcd(int a, int b)
{
	if (b == 0)
		return a;
	return gcd(b, a % b);
}
int main()
{
	int n, m, N, M;
	scanf("%d%d", &n, &m);
	scanf("%s%s", str1, str2);
	ll gcdd = gcd(n, m);
	N = n / gcdd;
	M = m / gcdd;
	ll lcm = n*m / gcdd;
	int flag = 1;
	for (int i = 0, j = 0;str1[i] && str2[j];i += M, j += N)
		if (str1[i] != str2[j]) { flag = 0;break; }
	if (flag)printf("%d\n", lcm);
	else printf("-1\n");
	//system("pause");
}