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 Julia (0.5.0) (judge verdict: Time Limit Exceeded)

function Zalgo(s)
	N=length(s)
	ret=zeros(Int,N)
	ret[1]=N
	i=1
	j=0
	while i<N
		while i+j<N&&s[j+1]==s[i+j+1]
			j+=1
		end
		ret[i+1]=j
		if j==0
			i+=1
			continue
		end
		k=1
		while i+k<N&&k+ret[k+1]<j
			ret[i+k+1]=ret[k+1]
			k+=1
		end
		i+=k
		j-=k
	end
	ret
end
const s=chomp(readline())
const t=chomp(readline())
const N=length(s)
const M=length(t)
function make()
now=s[:]
nowlen=N
while nowlen<M
	now*=s
	nowlen+=N
end
now
end
const now=make()
const AA=Zalgo(t*now*now)
const memo=-ones(Int,N)
const ok=[AA[i+M]>=M for i=1:N]
function dfs(u)
	if memo[u]>=0
		return memo[u]
	elseif memo[u]==-2
		return memo[u]=10^9
	end
	if ok[u]
		memo[u]=-2
		return memo[u]=dfs(mod1(u+M,N))+1
	else
		return memo[u]=0
	end
end
ans=0
for i=1:N
	global ans=max(ans,dfs(i))
end
println(ans<10^9?ans:-1)