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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,n) for(int i=(int)(j);i<(int)(n);i++)
#define REP(i,j,n) for(int i=(int)(j);i<=(int)(n);i++)
#define MOD 1000000007
#define int long long
#define ALL(a) (a).begin(),(a).end()
#define vi vector<int>
#define vii vector<vi>
#define pii pair<int,int>
#define priq priority_queue<int>
#define disup(A,key) distance(A.begin(),upper_bound(ALL(A),(int)(key)))
#define dislow(A,key) distance(A.begin(),lower_bound(ALL(A),(int)(key)))
#define tii tuple<int,int,int>
#define Priq priority_queue<int,vi,greater<int>>
#define pb push_back
#define mp make_pair
#define INF (1ll<<60)
signed main(){
  int N,M; cin>>N>>M;
  string S,T; cin>>S>>T;
  map<int,char> m;
  int X=N*M/__gcd(N,M);
  rep(i,0,N) m[X/N*i+1]=S[i];
  int ans=X;
  rep(i,0,M){
    int Y=X/M*i+1;
    if(m.count(Y)&&m[Y]!=T[i]){
      ans=-1;
      break;
    }
  }
  cout<<ans<<endl;
  }