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: Runtime Error)

/*
				   _ooOoo_
				  o8888888o
				  88" . "88
				  (| -_- |)
				  O\  =  /O
			   ____/`---'\____
			 .'  \\|     |//  `.
			/  \\|||  :  |||//  \
		   /  _||||| -:- |||||-  \
		   |   | \\\  -  /// |   |
		   | \_|  ''\---/''  |   |
		   \  .-\__  `-`  ___/-. /
		 ___`. .'  /--.--\  `. . __
	  ."" '<  `.___\_<|>_/___.'  >'"".
	 | | :  `- \`.;`\ _ /`;.`/ - ` : | |
	 \  \ `-.   \_ __\ /__ _/   .-` /  /
======`-.____`-.___\_____/___.-`____.-'======
				   `=---='
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
		fozubaoyou    pass System Test!
		quoted from "https://codeforces.com/contest/472/submission/7996840"
*/
 
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int,int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define fin(ans) cout<<(ans)<<endl
#define STI(s) atoi(s.c_str())
#define mp(p,q) make_pair(p,q)
#define pb(n) push_back(n)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define Sort(a) sort(a.begin(),a.end())
#define Rort(a) sort(a.rbegin(),a.rend())
#define PI acos(-1)
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
struct io{io(){ios::sync_with_stdio(false);cin.tie(0);};};
const int MOD = 1000000007;
const int INF = INT_MAX;
const ll LLINF = 1LL<<60;

ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a;}
ll lcm(ll a, ll b) { return a / gcd(a,   b) * b;}

int main(void){

	int n,m;
	string s,t;
	cin>>n>>m>>s>>t;
	int x=lcm(n,m);

	vector<char> u(x,'%');
	int pos=0;
	bool ch=1;
	for(int i=0;i<x;i+=x/n){
		u[i]=s[pos];
		pos++;
	}
	pos=0;
	// rep(i,u.size()){
	// 	cout<<u[i]<<" ";
	// }
	// cout<<endl;
	for(int i=0;i<x;i+=x/m){
		if(u[i]==t[pos]||u[i]=='%'){
			u[i]=t[pos];
		} else {
			ch=0;
		}
		pos++;
	}
	if(ch){
		fin(x);
	} else {
		fin(-1);
	}

}