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 (2.3.3) (judge verdict: Wrong Answer)

S = gets.chomp
T = gets.chomp

ss = S * (T.size / S.size.to_f).ceil

Ta = Array.new(T.size, 0)
Ta[0] = -1
i = 1
while i < T.size do
  if T[0] == T[i] then
    i, j = [i + 1, 1]
    while T[j] == T[i] && i < T.size do
      Ta[i] = j
      i, j = [i + 1, j + 1]
    end
    Ta[i] = j if i < T.size
  end
  i += 1
end

i, j, lst, sl, tl = [0, 0, [], ss.size, T.size]
while i < (3 * sl) do
  if ss[i % sl] == T[j] then
    if j == tl - 1 then
      lst << [i - j, i]
      j = Ta[j]
      i -= 1
    else
      j += 1
    end
  else
    if Ta[j] >= 0 then
      j = Ta[j]
      i -= 1
    end
  end
  i += 1
  break if 3 * sl - i - (tl - j) < 0
end

idx = 0
while idx < lst.size do
  f, t = lst[idx]
  a = lst.select{|ff, tt| ff == t + 1}
  if not a.empty? then
    lst[idx][1] = a[0][1]
    lst.delete_if{|ff, tt| f < ff && ff < a[0][1]}
    idx -= 1
  end
  idx += 1
end

ans = 0
if lst.empty? then
  ans = 0
elsif lst.size == 1 then
  ans = -1
else
  lst.each{|f, t|
    tmp = (t - f + 1) / tl
    ans = tmp if ans < tmp
  }
end

puts ans