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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
int main(void)
{

    int N;
    int M;

    scanf("%d %d", &N, &M);

    if(N < 1 || N > 100000 || M < 1 || M > 100000) {
        exit(0);
    }

    char S[N];
    char T[M];

    scanf("%s", S);
    scanf("%s", T);

    for(int i = 0;i < N;i++) {
        if((S[i] < 'a') || (S[i] > 'z')) {
            exit(0);
        }
    }

    for(int i = 0;i < M;i++) {
        if((T[i] < 'a') || (T[i] > 'z')) {
            exit(0);
        }
    }

    if(strlen(S) != N) {
        exit(0);
    }

    if(strlen(T) != M) {
        exit(0);
    }

    int hogeN = N;
    int hogeM = M;
    int tmp;

    if(hogeM < hogeN) {
        tmp = hogeM;
        hogeM = hogeN;
        hogeN = tmp;
    }

    int r;

    /* ユークリッドの互除法 */
    r = hogeM % hogeN;
    while(r != 0) {
        hogeM = hogeN;
        hogeN = r;
        r = hogeM % hogeN;
    }

    int saidai;
    int hoge;
    saidai = hogeN;

    int count = 1;
    int flag = 0;
    while(count / N < 1 || count / M < 1) {
        if(count % saidai != 0) {
            count++;
            continue;
        }
        hoge = count / N;
        for(int i = 1;i < N;i++) {
            if(S[i * hoge + i] != S[i]) {
                break;
            }
        }
        if(flag == 0) {
            count++;
            continue;
        }

        hoge = count / M;
        for(int i = 1;i < M;i++) {
            if(S[i * hoge + i] != S[i]) {
                break;
            }
        }

        if(flag == 0) {
            count++;
            continue;
        } else {
            flag = 1;
            break;
        }
    }

    if(flag == 1) {
        printf("%d", count);
    } else {
        printf("-1\n");
    }

    return 0;
}