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<bits/stdc++.h>
using namespace std;
#define ll long long
#define MP make_pair
#define pb push_back
#define F first
#define S second

const ll MX=1e5+100, mod=1e9+7;
int n, arr[MX];
ll fac=1, cnt, sum;

ll pn(ll a, ll b){
	if(b==0)return 1;
	ll nsf=pn(a, b/2);
	nsf*=nsf;
	nsf%=mod;
	if(b&1)nsf*=a;
	return nsf%mod;
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=0; i<n; i++){
		cin>>arr[i];
	}
	for(int i=1; i<=n; i++){
		fac*=i;
		fac%=mod;
	}
	for(int i=0; i<n; i++){
		cnt+=(ll)fac*pn(i+1, mod-2)%mod;
		cnt%=mod;
	}
	sum+=cnt*arr[0]%mod;
	for(int i=1; i<n; i++){
		cnt+=(ll)fac*pn(i+1, mod-2);
		cnt%=mod;
		cnt-=(ll)fac*pn(n-i+1, mod-2)%mod;
		cnt+=(ll)mod;
		cnt%=(ll)mod;
		sum+=(ll)cnt*arr[i]%mod;
		sum%=(ll)mod;
	}
	cout<<sum<<endl;
	return 0;
}