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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int get_lcm(int a, int b);
int get_gcd(int x, int y);

int main(void)
{
	int i, n, m, lcm, gcd;
	scanf("%d %d%*c", &n, &m);
	lcm = get_lcm(n,m);
	gcd = get_gcd(n,m);

	char s[100001], t[100001];
	for(i=0;i<n;i++)
		scanf("%c", &s[i]);
	s[i] = '\0';
	//printf("%s\n", s);
	scanf("%*c");
	for(i=0;i<m;i++)
		scanf("%c", &t[i]);
	t[i] = '\0';
	//printf("%s\n", t);

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

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

	return 0;
}

int get_lcm(int a, int b)
{
	int g, l, lcm;
	if(a>b) {
		g = a; l = b;
	} else {
		g = b; l = a;
	}
	lcm = g;

	while(lcm % l != 0) lcm += g;

	return lcm;
}

int get_gcd(int x, int y)
{
    int r;

    while((r = x % y) != 0)  // yで割り切れるまでループ
    {
        x = y;
        y = r;
    }
    return y;
}