Problem:
Score : 600 points

Problem Statement

Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.

There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.

Notes

A string a is a substring of another string b if and only if there exists an integer x (0 \leq x \leq |b| - |a|) such that, for any y (1 \leq y \leq |a|), a_y = b_{x+y} holds.

We assume that the concatenation of zero copies of any string is the empty string. From the definition above, the empty string is a substring of any string. Thus, for any two strings s and t, i = 0 satisfies the condition in the problem statement.

Constraints

1 \leq |s| \leq 5 \times 10^5

1 \leq |t| \leq 5 \times 10^5

s and t consist of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print -1.

Sample Input 1

abcabab
ab

Sample Output 1

3

The concatenation of three copies of t, ababab, is a substring of the concatenation of two copies of s, abcabababcabab, so i = 3 satisfies the condition.

On the other hand, the concatenation of four copies of t, abababab, is not a substring of the concatenation of any number of copies of s, so i = 4 does not satisfy the condition.

Similarly, any integer greater than 4 does not satisfy the condition, either. Thus, the number of non-negative integers i satisfying the condition is finite, and the maximum value of such i is 3.

Sample Input 2

aa
aaaaaaa

Sample Output 2

-1

For any non-negative integer i, the concatenation of i copies of t is a substring of the concatenation of 4i copies of s. Thus, there are infinitely many non-negative integers i that satisfy the condition.

Sample Input 3

aba
baaab

Sample Output 3

0

As stated in Notes, i = 0 always satisfies the condition.

Solution in Perl (5.26.1) (judge verdict: Time Limit Exceeded)

use strict;
use warnings;
package ZAlgorithm;
sub z_algorithm(@){
	my$n=@_;
	my@z=($n);
	my$i=1;
	my$j=0;
	while($i<$n){
		$j++while$i+$j<$n&&$_[$j]eq$_[$i+$j];
		$z[$i]=$j;
		if($j==0){
			$i++;
			next;
		}
		my$k=1;
		while($i+$k<$n&&$k+$z[$k]<$j){
			$z[$i+$k]=$z[$k];
			$k++;
		}
		$i+=$k;
		$j-=$k;
	}
	return@z;
}
package main;
my($s,$t)=map{chomp;$_}<>;
my@s=split"",$s;
my@t=split"",$t;
my@ss;
push@ss,@s while@ss<@t;
my@z=ZAlgorithm::z_algorithm(@t,@ss,@ss);
my@ok=map$z[$_+@t]>=@t,0..$#s;
local@main::memo=(-1)x@s;
sub dfs{
	my$u=shift;
	if($main::memo[$u]>=0){
		return$main::memo[$u];
	}
	elsif($main::memo[$u]==-2){
		return$main::memo[$u]=1e9;
	}
	if($ok[$u]){
		$main::memo[$u]=-1;
		return$main::memo[$u]=dfs(($u+@t)%@s)+1;
	}
	else{
		return$main::memo[$u]=0;
	}
}
my$ans=0;
for(0..$#s){
	my$now=dfs($_);
	$ans=$now if$ans<$now;
}
print$ans<1e9?$ans:-1,$/;