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>
#include<string.h>
void qz(long long a,long long b,long long*bs,long long*ys)
{
	int r,t,j;
	j=a*b;
	if(a<b)
	{
		t=a;
		a=b;
		b=t;
	}
	while(b!=0)
	{
		r=a%b;
		a=b;
		b=r;
	}
	*bs=j/a;
	*ys=a; 
}
int main()
{
	char s[100001],t[100001];
	int i;
	long long bs,ys,n,m;
	scanf("%ld",&n);
	scanf("%ld",&m);
	getchar();
	gets(s);
	gets(t);
	if(s[0]!=t[0])
	{
		printf("-1\n");
		return 0;
	}
	qz(n,m,&bs,&ys);
	if(n>m&&n%m==0)
	{
		for(i=1;i<m;i++)
		{
			if(s[m/n*i]!=t[i])
			printf("-1\n");
			return 0;
		}
	}
	else if(m>n&&m%n==0)
	{
		for(i=1;i<m;i++)
		{
			if(t[m/n*i]!=s[i])
			printf("-1\n");
			return 0;
		}
	}
	for(i=1;i<=ys;i++)
	{
		if(s[bs/m*i]!=t[bs/n*i])
		{
			printf("-1\n");
			return 0;
		}
	}
		printf("%ld",bs);
	return 0;
	 
}