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: Runtime Error)

#include <iostream>
#include<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<queue>
#include<regex>
#include<stack>
#include<stdio.h>
#include<vector>
#include<set>
#include<map>
#include<iomanip>
#define rep(i,n) for(int i=0;i<n;i++)
typedef int long long ll;
using namespace std;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};



int main(){
  
    int n,m;
    cin>>n>>m;
    int nn=n;int mm=m;
    string s,t;
    cin>>s>>t;
    
    if(s[0]!=t[0]){cout<<-1<<endl;return 0;}
    
    int g=0;
    
    if(mm>nn){swap(mm,nn);}
    while(true){
        g=nn%mm;
        if(g==0)break;
        nn=mm;
        mm=g;
    }
    
    mm=(n*m)/mm;
    
    int k=mm/n;
    int h=mm/m;
    
    char c[mm+1];
    
    rep(i,mm+1){
        c[i]='A';
    }
    
    rep(i,n){
        if(i*k>mm){cout<<-1<<endl;return 0;}
        c[i*k]=s[i];
    }
    
    rep(i,m){
        if(i*h>mm){cout<<-1<<endl;return 0;}
        if(c[i*h]!='A'&&c[i*h]!=t[i]){
            cout<<-1<<endl;return 0;
        }
    }
    
    cout<<mm<<endl;
return 0;
}