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: Accepted)

//#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <limits>
#include <map>
#include <vector>
#include <queue> 
#define P pair<int ,int >
#define pb push_back
#define LL long long
#define ft first
#define sd second
#define mp(x,y) make_pair(x,y)
#define db double
#define int LL
using namespace std;
const int N   = 1e5+233;
//const int mod = ;
const int INF =numeric_limits<int >::max();

#define rep(i,x,y) for (int i=x;i<=y;++i)
void read(int &x)
{
	x=0;
	char ch=getchar();
	int f=1;
	while (!isdigit(ch)) (ch=='-'?f=-1:0),ch=getchar();
	while ( isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	x*=f;
}

void write(int x)
{
	if (x<0) putchar('-'),x=-x;
	if (x>9) write(x/10);
	putchar(x%10+'0');
}
int gcd(int a,int b)
{
	return b ?gcd( b,a%b) :a;
}

int n,m;
char s[N],t[N];
signed main()
{
	read(n);read(m);
	scanf("%s%s",s,t);
	int tmp=gcd(n,m);
	/*if (s[1]!=t[1])
	{
		printf("-1");
		return 0;
	}*/
	if (tmp==n)
	{
		int ok=1;
		rep(i,0,m-1) if (s[i]!=t[i%n]) ok=0;
		printf("%lld",ok?m:-1);
		return 0;
	}
	if (tmp==m)
	{
		int ok=1;
		rep(i,0,n-1) if (s[i%m]!=t[i]) ok=0;
		printf("%lld",ok?n:-1);
		return 0;
	}
	int x=n/tmp,y=m/tmp;
	int ok=1;
	for (int i=0,j=0;i<n&&j<m&&ok;i+=x,j+=y) 
	  if (s[i]!=t[j]) 
	    ok=0;
	printf("%lld",ok? n*m/tmp :-1);
	return 0;
}