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<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<map>
#include<cstdio>
#include<stack>
#include<queue>
#include<fstream>
#include<cstdio>
#include<deque>
#include<iomanip>
#include <numeric>
//#include<bits/stdc++.h>

using namespace std;
#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000000000000000LL
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);(i)--)
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb push_back
#define pf push_front
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()
const double pi = acos(-1.0);

typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef deque<ll> dll;
typedef pair<ll, ll> P;
#define mod 1e9 + 7;
ll gcd(ll x, ll y)
{
	ll r;
	while ((r = x % y) != 0)
	{
		x = y;
		y = r;
	}
	return y;
}
ll lcm(ll x, ll y)
{
	x /= gcd(x, y);
	y /= gcd(x, y);
	return (x * y);
}


int main() {
	ll n, m;
	string s, t;
	cin >> n >> m;
	cin >> s;
	cin >> t;

	ll lc = lcm(n, m);
	bool isOK = true;
	ll L = lc;
	ll ln = L / n;
	ll lm = L / m;

	int j = 0;
	for (int i = 0; i < n / lm; i++) {
		if (s[i*lm] != t[i*ln]) {
			cout << "-1" << endl;
			return 0;
		}
	}


	cout << lcm << endl;

	return 0;
}