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 <algorithm>
#include <string>
#include <vector>
#include <math.h>
#define MOD 1000000007
typedef long long ll;
using namespace std;

int gcd(int x,int y){
  return y==0?x:gcd(y,x%y);
}

int main(){
  int n,m;
  cin>>n>>m;
  string s,t;
  cin>>s;
  cin>>t;

  if(s[0]!=t[0]){
    cout<<-1<<endl;
    return 0;
  }

  int z=gcd(n,m);
  int a=n/z;
  int b=m/z;
  int lcm=z*a*b;

  bool flag=0;
  for(int i=0;i*a<n&&i*b<m;i++){
    if(s[i*a]!=t[i*b]) flag=1;
  }
  if(flag) cout<<-1<<endl;
  else cout<<lcm<<endl;

  /*if(n==m){
    if(s==t) cout<<n<<endl;
    else cout<<-1<<endl;
  }else if(n<m){
    if(z==1) cout<<lcm<<endl;
    else{
      bool flag=0;
      for(int i=0;i<n;i++){
        if(s[i]!=t[i*b]) flag=1;
      }
      if(flag) cout<<-1<<endl;
      else cout<<lcm<<endl;
    }
  }else{
    if(z==1) cout<<lcm<<endl;
    else{
      bool flag=0;
      for(int i=0;i<m;i++){
        if(t[i]!=s[i*a]) flag=1;
      }
      if(flag) cout<<-1<<endl;
      else cout<<lcm<<endl;
    }
  }*/


  return 0;
}