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: Accepted)

#include<iostream>
#include<climits>
#include<vector>
#include<list>
#include<functional>
#include<algorithm>
#include<string>
#include<cmath>
#include<complex>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) for(int i=0;i<(n);i++)
#define INF (int)1e9
#define INFL (ll)1e11
using namespace std;
typedef long long ll;
typedef pair<int, int> P;

int gcd(int x, int y) {
	return y == 0 ? x : gcd(y, x%y);
}

ll lcm(int x, int y) {
	return (ll)x*y / gcd(x, y);
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N, M;
	string S, T;
	cin >> N >> M >> S >> T;
	
	int g = gcd(N, M);
	ll l = lcm(N, M);
	int a = l / M, b = l / N;
	REP(i, g) {
		if (S[i*a] != T[i*b]) {
			l = -1;
			break;
		}
	}
	cout << l << "\n";

	return 0;
}