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++ (Clang 3.8.0) (judge verdict: Runtime Error)

#include <iostream>
#include <sstream>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <tuple>
#include <cassert>

using namespace std;

#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
#define pb(e) push_back(e)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define mp(a,b) make_pair((a), (b))
#define mt(a,b,c) make_tuple((a),(b),(c))

typedef long long ll;
long long int gcd(long long int x,long long int y){
    long long int r;
    if(x == 0 || y == 0)  return 0;

    while((r = x % y) != 0) 
    {
        x = y;
        y = r;
    }
    return y;
}

int main(){
    long long int n,m;
    string s,t; cin >> n >> m >> s >> t;
    if(n==m){
      assert(0);
        if(s == t) cout << n << endl;
        else cout << -1 << endl;
        return 0;
    }
    if(n<m){swap(s,t); swap(n,m);}
    ll lcm = (n * m) / gcd(n,m);
    if(lcm == n){
        rep(i,m){
            if(s[(lcm / m) * i] != t[i]){
                cout << -1 << endl;
                return 0;
            }
        }
    }
    if(s[0] == t[0]){cout << lcm << endl;}
    else cout << -1 << endl;
    return 0;
}