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 <bits/stdc++.h>
#define  rep(i,a,b)for(int i=a;i<b;++i)
#define  rrep(i,a,b)for(int i=a;i>=b;--i)
#define  yesno(flag)if(flag)cout<<"Yes"<<endl; else cout<<"No"<<endl;
using namespace std;
typedef long long ll;
const int INF=1000000000LL;
const ll r197=1000000007;
typedef pair<int,int>P;
typedef pair<int,P>Pa;
char e[100001],f[100001];

int main() {
	ll n,m;
	string s,t;
	cin>>n>>m;
	cin>>s>>t;
	ll a=n,b=m;
	while(1){
		if(a<b){
			ll d=a;
			a=b;
			b=d;
		}
		if(!b)break;
		a%=b;
	}
	ll l=(n*m)/a;
	rep(i,0,n){
		e[i]=s[i*(l/n)];
	}
	rep(i,0,m){
		f[i]=t[i*(l/m)];
	}
	bool ans=false;
	ll x=0,y=0;
	while(1){
		if(x>=n||y>=m)break;
		if(x*l/n==y*l/m){
			if(e[x]!=f[y])ans=true;
			if(x<n-1)x++;
			else y++;
		}else{
			if(x*(l/n)>y*(l/m))y++;
			else if(x*(l/n)<y*(l/m))x++;
		}
	}
	if(!ans)cout<<l<<endl;
	else cout<<-1<<endl;
	return 0;
}