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 Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

# frozen_string_literal: true

def z_algorithm(s)
  n = s.size
  return [] if n == 0

  z = [0]*n
  z[0] = n
  i, j = 1, 0
  while i < n
    j += 1 while i+j<n && s[j]==s[i+j]
    z[i] = j
    if j == 0 then
      i += 1
      next
    end
    k = 1
    while i+k<n && k+z[k]<j
      z[i+k] = z[k]
      k += 1
    end
    i += k
    j -= k
  end
  return z
end

s, t = $<.map &:chomp

ss = s+s
ss += s while ss.size < t.size*2
z = z_algorithm(t+ss)[t.size,s.size].map{ |z| z >= t.size }

vis = [false]*s.size

ans = (0 ... s.size).map{ |u|
  next if vis[u]
  vis[u] = true
  now = u
  cnt = 0
  while z[now]
    now = (now+t.size) % s.size
    vis[now] = true
    cnt += 1
    break if now == u
  end
  now==u && cnt.nonzero? ? 10**9 : cnt
}.compact.max

ans = -1 if ans >= 10**9
puts ans