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 <bits/stdc++.h>
using namespace std;
const int MAXN = 100010;
typedef long long LL;
char s[MAXN], t[MAXN];


int main() {
	int n, m; bool flag = true;
	scanf("%d %d", &n, &m);
	scanf("%s %s", s, t);
	int ans = __gcd(n, m);
	for(int i = 0; ; ++i) {
		int cnt1 = n / ans * i;
		int cnt2 = m / ans * i;
		//printf("%d %d %d\n", i, cnt1, cnt2);
		if(cnt1 >= n || cnt2 >= m) break;
		//puts("$$");
		if(s[cnt1] != t[cnt2]) {
			flag = false;
			break;
		}
	}
	if(flag) printf("%d\n", n * m / ans);
	else puts("-1");
	return 0;
}