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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
const ull mod = 1e9 + 7;
#define REP(i,n) for(int i=0;i<(int)n;++i)

ull Euclid(ll A, ll B){
    if(A<B) swap(A, B);
    ull C = B;
    while(C != 0){
        C = A%B;
        A = B;
        B = C;
    }
    return A;
}

int main(){
	ll N, M;
	string S, T;
	cin >> N >> M;
	cin >> S;
	cin >> T;
	if(N < M){
		swap(N, M);
		swap(S, T);
	}
	ll E = Euclid(N, M);
	ll kou = E * (N/E) * (M/E);
	bool flag = true;
	vector<int> res(kou);
	ll a = kou/N;
	REP(i, kou){
		if(i%a==0){
			res[i] = (int)S[i/a] - 97;
		}else{
			res[i] = -1;
		}
	}
	/*
	REP(i, kou) cout << res[i] << " ";
	cout << endl;
	*/
	ll b = kou/M;
	REP(i, M){
		if(res[b*i]!=(int)T[i] - 97 && res[b*i]!=-1){
			flag = false;
		}
	}
	if(flag){
		cout << kou << endl;
	}else{
		cout << -1 << endl;
	}
    return 0;
}