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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define MOD 1000000007
using namespace std;
long long int gcd(long long int a,long long int b)
{
    while(a % b != 0)
    {
        int r = a % b;
        a = b;
        b = r;
    }
    return b;
}
int main()
{
    long long int n,m;
    string s,t;
    cin >> n >> m >> s >> t;
    long long int l = n * m / gcd(m,n);
    vector<pair<int,char> > v1,v2;
    int i = 0,j = 0;;
    while(i < l)
    {
        v1.push_back(make_pair(i,s[j]));
        i += l / n;
        j++;
    }
    i = 0,j = 0;
    while(i < l)
    {
        v2.push_back(make_pair(i,t[j]));
        i += l / m;
        j++;
    }
    bool flag = true;
    for(int i = 0;i < v1.size();i++)
    {
        for(int j = 0;j < v2.size();j++)
        {
            if(v1[i].first == v2[j].first && v1[i].second != v2[j].second)
            {
                flag = false;
                break;
            }
        }
    }
    if(flag)
    {
        cout << l << "\n";
    }
    else
    {
        cout << "-1\n";
    }

    return 0;
}