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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdint.h>
#include<stdbool.h>
#include<tgmath.h>
#include<sys/queue.h>
#include<time.h>
#define ll long long
#define min(a, b) ((a) > (b) ? (b) : (a))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define miin(a, b, c) min((a), min((b), (c)))
#define maax(a, b, c) max((a), max((b), (c)))
#define mod 1000000007
#define INF 1001001001
#define NIL -1

int n, m;
char s[111111], t[111111];

ll gcd(ll a, ll b)
{
    return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}

main(){
    scanf("%d%d", &n, &m);
    scanf("%s%s", s, t);
    bool flag = true;
    int i=0, j=0, d = gcd(n, m);
    while(i < n){
        if(s[i] != t[j])flag = false;
        i += n/d;
        j += m/d;
    }
    printf("%lld\n", flag?lcm(n, m):-1);
    return 0;
}