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;
int n,m,l=0,lmax;
char s[100005],t[100005],x[100005];
int gb(int a,int b)
{
	int m=a,n=b,c,d;
	while(1){//循环的辗转相除法 
		c=a%b;
		a=b;
		b=c;
		if(b==0){
			break;
		}
	}
	c=m/a;d=n/a;
	n=c*d*a;
	return n;
}
int main()
{
	cin>>n>>m;
	l=gb(n,m);
	int i,j;
	for(i=1;i<=100004;i++)
		x[i]='0';
	int flag=0,result=0;
	scanf("%s",s+1);
	scanf("%s",t+1);
	if(s[1]!=t[1]){
		cout<<"-1";
		return 0;
	}
	if(n>=10000)
	{
		cout<<l;
		return 0;
	}
	int k1=l/n,k2=l/m;
	j=1;
	int sn=1,tn=1;
	for(i=1;i<=l;i+=k1,sn++)
	{
		if(flag)
			break;
		for(;j<=l;j+=k2,tn++)
		{
			if(j>i)
				break;
			if(j<i)
				continue;
			if(j==i)
			{
				if(s[sn]!=t[tn])
				{
					flag=1;
					cout<<"-1";
					break;
				}
			}
		}
	}
	if(flag==0)
		cout<<l;
 }