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: Wrong Answer)

#include <iostream>
#include <string>
using namespace std;

int main() {
	int n,m;
	string s,t;
	cin >> n >> m;
	cin >> s;
	cin >> t;

  int x = n * m;
 
  /* 自然数 n > m を確認・入替 */
  if(n<m){
    int tmp = n;
		string tmpstr = s;
    n = m;
		s = t;
    m = tmp;
		t = tmpstr;
  }
  /* ユークリッドの互除法 */
  int r = n % m;
	int gcd = m;
  while(r!=0){
		int n1 = gcd;
		gcd = r;
    r = n1 % gcd;
  }
  /* 最小公倍数 */
	int lcm = x/gcd;

/*	if (s[0]!=t[0]) cout << -1 << endl;
	else if (n==m&&s!=t) cout << -1 << endl;
	else if (gcd>1) {*/
		for (int i=0; i<=gcd-1; i++) {
			if (s[n/gcd*i]!=t[m/gcd*i]) {
				cout << -1 << endl;
				break;
			}
			if (i==gcd-1) {cout << lcm << endl; return 0;}
		}
	/*} else*/ cout << lcm << endl;

	return 0;
}