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 <iostream>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cstdlib>
#include <stdio.h>
#include <queue>
#include <stack>
#include <string>
#include <cmath>
#include <vector>
#include <map>
using namespace std;
int gcd(int x,int y){
	if(y==0) return x;
	return gcd(y,x%y);
} 
int lcm(int x,int y){
	return x*y/gcd(x,y);
}
map<long long,long long> a;
map<long long,long long> b;
map<long long,long long> c;
map<long long,long long> vis;
char q[1000002];
char w[1000002];
int main(){	
	int n,m;
	scanf("%d%d",&n,&m);
	scanf("%s",q);
	scanf("%s",w);
	for(int i=0;i<n;i++) a[i]=q[i];
	for(int i=0;i<m;i++) b[i]=w[i];
	int p=0;
	int ll=lcm(n,m);
		if(n>m)
			p=1;
		else p=0;
		for(int j=0;j<max(n,m);j++){
			if((vis[j*(ll/m)]&&c[j*(ll/m)]!=b[j])||(vis[j*(ll/n)]&&c[j*(ll/n)]!=a[j])){
				cout<<"-1";
				return 0;
			}
			if(j*(ll/m)==j*(ll/n)&&a[j]!=b[j]){
				cout<<"-1";
				return 0;
			}
			if(j==m-1&&p==1){
				cout<<ll;
				return 0;
			}
			if(j==n-1&&p==0){
				cout<<ll;
				return 0;
			}
			c[j*(ll/m)]=b[j],c[j*(ll/n)]=a[j];
			vis[j*(ll/m)]=1;
			vis[j*(ll/n)]=1;
		}
		cout<<ll;
		return 0;
	return 0;
}