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: Wrong Answer)

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
// #include<ext/pb_ds/assoc_container.hpp>
// #include<ext/pb_ds/tree_policy.hpp>
// #include<ext/pb_ds/tag_and_trait.hpp>
// using namespace __gnu_pbds;
// #include<boost/multiprecision/cpp_int.hpp>
// namespace multiprecisioninteger = boost::multiprecision;
// using cint=multiprecisioninteger::cpp_int;
using namespace std;
using ll=long long;
#define double long double
using datas=pair<ll,ll>;
using ddatas=pair<double,double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
// using llset=tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;
#define For(i,a,b) for(i=a;i<(ll)b;++i)
#define bFor(i,b,a) for(i=b,--i;i>=(ll)a;--i)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N,0)
#define brep1(i,N) bFor(i,N,1)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define endl "\n"
#define eb emplace_back
#define print(v) cout<<v<<endl
#define printyes cout<<"Yes"<<endl
#define printno cout<<"No"<<endl
#define printYES cout<<"YES"<<endl
#define printNO cout<<"NO"<<endl
#define output(v) do{bool ff=0;for(auto outi:v){cout<<(ff?" ":"")<<outi;ff=1;}cout<<endl;}while(0)
#define matoutput(v) do{for(auto outimat:v)output(outimat);}while(0)
const ll mod=1000000007;
// const ll mod=998244353;
const ll inf=1LL<<60;
const double PI = acos(-1);
const double eps = 1e-9;
template<class T> inline bool chmax(T& a,T b){bool x=a<b;if(x)a=b;return x;} 
template<class T> inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;} 

void startupcpp(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(15);
}
int v[1500001];
char s[1500001],t[500002];
void Zalgorithm(int N){
  int i=1,j=0,k=0;
  while(i<N){
    while(i+j<N&&s[i+j]==s[j])++j;
    v[i]=j;
    if(!j){
      ++i;
      continue;
    }
    k=1;
    while(v[k]+k<j){
      v[i+k]=v[k];
      ++k;
    }
    i+=k;
    j-=k;
  }
}
int i,j,K,quef[1000001],ques[1000001],size,ans;
int main(){
  int N,ng=500001,mid;
  fgets(t,1000000,stdin);
  fgets(s,1000000,stdin);
  while(ng-K>1){
    mid=(ng+K)>>1;
    (s[mid]?K:ng)=mid;
  }
  while(t[i]!='\n'){
    s[K+i]=t[i];++i;
  }
  N=i;
  while(i<1000000){
    s[K+ i++]=t[j++];
    j*=j<N;
  }
  N=(K+N-1)/N*N;
  Zalgorithm(1500000);
  rep(i,N){
    if(v[i+K]<K)ques[size++]=i+K-(i+K>=N)*N;
  }
  rep(i,N){
    if(i==size){
      puts("-1");
      return 0;
    }
    chmax(ans,quef[i]);
    if(v[ques[i]+=K]>=K){
      ques[size]=ques[i]-N*(ques[i]>=N);
      quef[size++]=quef[i]+1;
    }
  }
  printf("%d\n",ans);
}