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 LuaJIT (2.0.4) (judge verdict: Wrong Answer)

local s = io.read()
local t = io.read()
local swapped = false
if #s < #t then
  swapped = true
  s, t = t, s
end
local slen = #s
local tlen = #t
s = s .. s

local mma = math.max
local findpos = s:find(t, 1, true)
local maxlength = 0
local mfl = math.floor
while findpos do
  if slen < findpos then
    break
  end
  local curlength = 1
  while true do
    local p = findpos + tlen * curlength
    if p < findpos + slen and t == s:sub(p, p + tlen - 1) then
      curlength = curlength + 1
    else
      break
    end
  end
  maxlength = mma(maxlength, curlength)
  if slen % tlen == 0 and curlength == mfl(slen / tlen) then
    maxlength = -1
    break
  end
  findpos = s:find(t, findpos + tlen * curlength, true)
end

local same_s, same_t = true, true
for i = 2, slen do if s:sub(1, 1) ~= s:sub(i, i) then same_s = false break end end
for i = 2, tlen do if t:sub(1, 1) ~= t:sub(i, i) then same_t = false break end end
if same_s and same_t and s:sub(1, 1) == t:sub(1, 1) then
  maxlength = -1
end
if swapped then
  if 0 < maxlength then
    maxlength = 0
  end
end
print(maxlength)