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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a,long long b){
	if(b==0) return a;
	return gcd(b,a%b);
}
int main(){
	long long n,m;
	string s,t;
	cin>>n>>m;
	cin>>s;
	cin>>t;
//	string ans;
	long long G;
	if(n>m){
		G=n*m/gcd(n,m);
	}else{
		G=n*m/gcd(m,n);
	}
	if(s[0]!=t[0]){
		cout<<"-1"<<endl;
	}else{
		bool flag=false;
		int aans;
		for(int i=1;i<=100005;i++){
			flag=true;
			long long x=G*i/n;
			long long y=G*i/m;
			int k,j;
			k=j=0;
			while(k!=n&&j!=m){
				long long m;
				if(x*k+1<y*j+1){
					m=x*k+1;
					k++;
				}else if(x*k+1==y*j+1){
					m=x*k+1;
					if(s[k]!=t[j]){
						flag=false;
					}
					k++;
					j++;
				}else{
					m=y*j+1;
					j++;
				}
			}
			if(flag){
				aans=i;
				break;
			}
		}
		if(flag){
			long long res=G*aans;
			cout<<res<<endl;
		}else{
			cout<<"-1"<<endl;
		}
	}
	return 0;
}