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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

//------------------------------>> tsukasa_diary's template <<------------------------------//
#include <bits/stdc++.h>
using namespace std;

#define for_(i,a,b) for(int i=(a);i<(b);++i)
#define allof(a) (a).begin(),(a).end()
using lint = long long;
template< typename T > using Vec = vector< T >;

template< typename T > bool inRange(T x, T lb, T ub) { return lb <= x && x < ub; }
template< typename T > bool inRange(T x, T y, T lb, T ub) { return in_range(x, lb, ub) && in_range(y, lb, ub); }
template< typename T > void modAdd(T& a, T b, T mod) { a = (a + b + mod) % mod; }
template< typename T > void modMul(T& a, T b, T mod) { a = (a * b) % mod; }
template< typename T > void minUpdate(T& a, T b) { a = min(a, b); }
template< typename T > void maxUpdate(T& a, T b) { a = max(a, b); }
inline int bitCount(int x) { return __builtin_popcount(x); }
inline int bitCount(lint x) { return __builtin_popcountll(x); }

const int dx[4] = {0,1,0,-1}, dy[4] = {-1,0,1,0};
const int Dx[8] = {0,1,1,1,0,-1,-1,-1}, Dy[8] = {-1,-1,0,1,1,1,0,-1};
const double PI = acos(-1);

//--------------8---------------->> coding space <<-----------------8-------------//

Vec< int > makeKMPfail(const string& S) {
	int n = S.size();	
	Vec< int > fail(n+1, 0);
	int j = fail[0] = -1;
	for_(i,1,n+1) {
		while (j >= 0 && S[j] != S[i - 1]) j = fail[j];
		fail[i] = ++j;
	}
	return fail;
}

bool cyclic(const string& S, const string& T) {
	string x = S + S;
	Vec< int > t_fail = makeKMPfail(T);
	int N = x.size(), m = T.size();
	for (int i=0, k=0; i < N; ++i) {
		while (k >= 0 && T[k] != x[i]) k = t_fail[k];
		if (++k >= m) return true;
	}
	return false;
}

int main() {
	string s, t;
	cin >> s >> t;
	int n = s.size(), m = t.size();
	
	///
	Vec< int > s_fail = makeKMPfail(s);
	Vec< int > t_fail = makeKMPfail(t);
	int sp = n - s_fail[n], tp = m - t_fail[m];
	if (n % sp == 0 && m % tp == 0) {
		if (sp == tp && cyclic(s.substr(0, sp), t.substr(0, tp))) {
			puts("-1");
			return 0;
		}
	}
//	cerr << "sp " << sp << endl;
//	cerr << "tp " << tp << endl;
	
	///
	string x = s;
	while (x.size() < t.size()) x += x;
	for_(i,0,5) x += x;
	int N = x.size();
//	cerr << x << endl;
//	cerr << t << endl;
	
	Vec< bool > match(N+1, false);
	for (int i=0, k=0; i < N; ++i) {
		while (k >= 0 && t[k] != x[i]) k = t_fail[k];
		if (++k >= m) {
			match[i] = true;
			k = t_fail[k];
		}
	}
	
	int ans = 0;
	for_(i,1,N+1) {
		if (match[i]) {
			int j = i, cnt = 0;
			while (j <= N && match[j]) {
				match[j] = false;
				j += m;
				++cnt;
			}
			ans = max(ans, cnt);
		}
	}
	cout << ans << endl;
}
//--------------8---------------->> coding space <<-----------------8-------------//