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: Runtime Error)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <iomanip>
#include <cstring>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n-1); i >= 0; i--)
#define pb push_back
#define f first
#define s second
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll max3(ll a, ll b, ll c){return max(a, max(b, c));};
ll min3(ll a, ll b, ll c){return min(a, min(b, c));};
ll max4(ll a, ll b, ll c, ll d){return max(max(a, b), min(c, d));};
ll min4(ll a, ll b, ll c, ll d){return min(min(a, b), min(c, d));};
ll max5(ll a, ll b, ll c, ll d, ll e){return max(max(a, b), max3(c, d, e));};
ll min5(ll a, ll b, ll c, ll d, ll e){return min(min(a, b), min3(c, d, e));};

const ll INFL = 1LL << 60;//10^18 = 2^60
const int INF = 1 << 30;//10^9
ll MOD = 1000000007;
//ll MOD = 998244353;
vector<ll> dy = {0, 0, 1, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, -1, 0, 0, 1, -1, 1, -1, 0};

ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}
ll pow_mod(ll x, ll k){
  x %= MOD; x += MOD; x %= MOD;
  ll res = 1;
  while(k > 0){
    if(k % 2){
      res *= x; res %= MOD;
    }
    x *= x; x %= MOD;
    k /= 2;
  }
  return res;
}

ll inverse(ll x){return pow_mod(x, MOD - 2);};

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y){return x / gcd(x, y) * y;};

ll kai_mod(ll x){
  if(x == 0) return 1;
  return x * kai_mod(x-1) % MOD;
}

/*
//コンビネーション
const int MAXcomb = 200010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i = 2; i < MAXcomb; i++){
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
ll comb(int n, int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
}
*/

string S, T;
vector<bool> canStart;
const ll MAX_V = 200010;

ll par[MAX_V];//
ll Rank[MAX_V];
ll child[MAX_V];

void init_UF(){
  for(ll i = 0; i < MAX_V; i++){
    par[i] = i;
    Rank[i] = 1;
    if(canStart[i]) child[i] = 1;
  }
}

ll find(ll x){
  if(par[x] == x) return x;
  else{
    return par[x] = find(par[x]);
  }
}

bool same(ll x, ll y){
  return find(x) == find(y);
}

void unite(ll x, ll y){
  ll rx = find(x), ry = find(y);
  if(rx != ry){
    if(Rank[rx] >= Rank[ry]){
      par[ry] = rx;
      if(Rank[rx] == Rank[ry]) Rank[rx]++;
      child[rx] += child[ry];
    }else{
      par[rx] = ry;
      child[ry] += child[rx];
    }
  }
}

int main(){
  cin >> S >> T;
  canStart.resize(S.size(), 0);
  ll a = 998244353;
  ll hashS = 0, hashT = 0;
  rep(i, T.size()){
    hashS = (hashS * a + S[i % S.size()]) % MOD;
  }
  rep(i, T.size()){
    hashT = (hashT * a + T[i]) % MOD;
  }
  ll po = pow_mod(a, T.size());
  rep(i, S.size()){
    if(hashS == hashT) canStart[i] = 1;
    hashS = ((hashS * a + S[(i + T.size()) % S.size()] - S[i] * po) % MOD + MOD) % MOD;
  }
  init_UF();
  bool inf = 0;
  rep(i, S.size()){
    ll to = (i + T.size()) % S.size();
    if(canStart[i] == 0 || canStart[to] == 0) continue;
    if(same(i, to)) inf = 1;
    unite(i, to);
  }
  ll ans = 0;
  if(inf) cout << -1 << endl;
  else{
    rep(i, S.size()){
      ans = max(ans, child[i]);
    }
    cout << ans << endl;
  }
}