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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;
public class Main {
    static int lcm(int n,int m){
        if(n==m) return n;
        int ans=1;
        int div=2;
        int max_kouyaku=Math.abs(m-n);
        while(div==2){
            if(n%2==0 && m%2==0){
                n/=2; m/=2;
                ans*=div;
            }
            else div++;
        }
        for(div=3;div<=Integer.min(n,m) && div<=max_kouyaku;div+=2){
            if(m%div==0 && n%div==0){
                m/=div; n/=div;
                ans*=div;
                div-=2;
            }
        }
        return ans*m*n;
    }
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int N=sc.nextInt();
        int M=sc.nextInt();
        int lcm=lcm(N,M);
        int stepS=lcm/N;
        int stepT=lcm/M; //stepS,stepT刻みで入る
        int dupS=lcm(stepS,stepT)/stepS;
        int dupT=lcm(stepS,stepT)/stepT;
        //文字が一致しなくてはならないのはSの1+n(dupS)番目の文字とTの1+n(dupT)番目の文字
        String S=sc.next();
        String T=sc.next();
        for(int i=0;i<N/dupS;i++){
            if(S.charAt(i*dupS)!=T.charAt(i*dupT)){
                System.out.println(-1);
                return;
            }
        }
        System.out.println(lcm);
    }
}