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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
#define ll long long
using namespace std;
map<ll,char>mp;
int main()
{
    ll n,m;
    string a,b;
    scanf("%lld%lld",&n,&m);
    cin>>a>>b;
    ll ans;
    if(a[0]!=b[0]) {printf("-1\n");return 0;}
    ll k=min(n,m);
    long long xxx=n*n*m*m;
    for(int j=1;j<=xxx;j++)
    {
        if(k*j%n==0&&k*j%m==0)
        {
            ans=k*j;
            bool flag=true;
            ll x=ans/n,y=ans/m;
            mp.clear();
            for(int i=1;i<n;i++)
            {
                ll tj=i*x;
                mp[tj]=a[i];
            }
            for(int i=1;i<m;i++)
            {
                ll tj=i*y;
                if(mp[tj]==0) continue;
                if(mp[tj]!=b[i]) {flag=false;break;}
            }
            if(flag)
            {
                printf("%lld\n",ans);return 0;
            }
        }
    }
    printf("-1\n");
}