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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <iterator>
#include <map>
#include <set>
#include <iomanip>
#include <vector>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using Graph = vector<vector<int>>;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
#define INF 100000000000
#define MAX 200001
#define MOD 1000000007

int gcd(int a, int b){
    if (a % b == 0){
        return (b);
    }
    else{
        return (gcd(b, a % b));
    }
}

int lcm(int a, int b)
{
    return a * b / gcd(a, b);
}

int main(){
    int N, M;
    cin >> N >> M;
    string s, t;
    cin >> s >> t;
    int G = lcm(N, M);
    int flag = 1;
    if(G == N or G == M){
        if(N < M){
            for(int i = 0; i < N; i++){
                if(s[i * (G/N)] != t[i * (G/M)]){
                    flag = 0;
                    break;
                }
            }
        }
        else{
            for (int i = 0; i < M; i++){
                if (s[i * (G/M)] != t[i * (G/M)]){
                    flag = 0;
                    break;
                }
            }
        }
    }
    else{
        if(s[0] != t[0]) flag = 0;
    }
    if(flag) cout << G << endl;
    else cout << -1 << endl;
}