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 <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <climits>
#include <string>
#include <map>
#include <set>
#include <list>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;A++)

using namespace std;

int i,j,k,l,m,n;
/////////////////////////////////////////////////////


ll a;
int N,M;
char S[101010],T[101010];
int gcd(int a,int b){
    if(b==0){
        return a;
    }
    return gcd(b,a%b);
}

int main(){
    
    scanf("%d%d%s%s",&N,&M,S,T);
    ll g=gcd(N,M);
    ll l=N*M/g;
    ll n=N/g,m=M/g;
    ll ni=0,mi=0;
    ll ans=0;
    for(i=0;i<l;i+=n*m){
        if(S[i/m]!=T[i/n]){
            printf("-1\n");
            return 0;
        }
    }

    printf("%d\n",l);
    return 0;
}