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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
long min_l(long n, long m);

int main(void){
	/* code */
	long n, m, max, min, i, k, p1, p2, q1, q2, answer;
	int err;
	char *s, *t;
	err=scanf("%ld%ld", &n, &m);
	if (err!=2) exit(1);
	s = (char *)malloc(sizeof(char)*(int)n);
	t = (char *)malloc(sizeof(char)*(int)m);
	err=scanf("%s\n%s", s, t);
	if (err!=2) exit(1);

	k=min_l(n, m);
	p1=k/n;
	q1=k/m;
	p2=0;
	q2=0;
	answer = k;
	while(q2<n&&p2<m){
		/* code */
		if (s[q2]!=t[p2]){
			answer = -1;
		}
		if(answer==-1) break;
		p2+=p1;
		q2+=q1;
	}
	printf("%ld\n", answer);
	free(s);
	free(t);
	return 0;
}

long min_l(long n, long m){
	long answer = 1, i = n;
	while (i >= 2)
	{
		if ((n%i==0) && (m%i==0))
		{
			answer *= i;
			n /= i;
			m /= i;
			i = n;
		}else
		{
			i--;
		}
	}
	answer = answer * n * m;
	return answer;
}