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<cstdio>
#include<algorithm>
long long int gcd(long long int a, long long int b){
	if(a<b){
		long long int c=a;
		a=b;
		b=c;
	}
	if(a%b==0){
		return b;
	}else{
		return gcd(b, a%b);
	}
}
int main(){
	long long int n,m;
	scanf("%lld%lld",&n,&m);
	char s[n]={};
	char t[m]={};
	scanf("%s%s",s,t);
	long long int l=n*m/gcd(n ,m);
	int mode=0;
	for(long long int i=0;i<gcd(n, m);i++){
		if(s[l*i/m]!=t[l*i/n]){
			mode=1;
		}
	}
	if(mode==0)printf("%lld\n",l);
	else printf("-1\n");
	return 0;
}