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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(void){
	int n,m;
	char s[100001],t[100001];
	scanf("%d %d",&n,&m);
	scanf("%s",s);
	scanf("%s",t);
	int a,b,c;
	if(n>m){
		a=n;
		b=m;
	}else{
		a=m;
		b=n;
	}
	while(1==1){
		c=a%b;
		if(c==0){
			break;
		}
		a=b;
		b=c;
	}
	int ans;
	ans = n*m/b;

	int i;
	for(i=0;i<b;i++){
		if(s[(n/b)*i]!=t[(m/b)*i]){
			ans = -1;
			break;
		}
	}
	printf("%d\n",ans);
	return 0;
}