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)

//Head File
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#include<ctime>
#include<set>
#include<cstdlib>
#include<vector>
#include<cassert>
#include<cmath>
#include<iostream>
using namespace std;

#define il inline

//Variable
#define ll long long
#define ull unsigned long long
#define db double
#define lb long double
#define ui unsigned int

//Debug
#define B cerr<<"Break Point"<<endl;
#define P(x) cerr<<#x<<' '<<"="<<' '<<(x)<<endl;
#define p(x) cerr<<#x<<' '<<"="<<' '<<(x)<<' ';
#define ml(x) cerr<<"Size of array is "<<x*4/1024/1024<<" MB"<<endl;

//Vector
#define vc vector
#define puf push_front
#define pof pop_front
#define pub push_back
#define pob pop_back
#define vbe(x) x.begin(),x.end()

//Memset
#define ms(x) memset(x,0,sizeof(x))
#define MS(x) memset(x,0x3f3f3f3f,sizeof(x))

//Pair
#define fi first
#define se second

//File

#define fin(x) freopen(x,"r",stdin)
#define fou(x) freopen(x,"w",stdout)
void fio()
{
    #ifndef ONLINE_JUDGE
    freopen("sample.in","r",stdin);
    freopen("sample.out","w",stdout);
    #endif
}
void pti()
{
    double timeuse = clock()*1000.0/CLOCKS_PER_SEC;
    cerr<<"Timeuse "<<timeuse<<"ms"<<endl;
}
void end()
{
    #ifndef ONLINE_JUDGE
    pti();
    #endif
    exit(0);
}

//Inf
#define INF 0x3f3f3f3f
#define LINF ((long long)(0x3f3f3f3f3f3f3f3f))

//IO
#define pc(s) putchar(s)
#define say(s) cout<<s<<endl
namespace io
{
    const int SIZ=55;int que[SIZ],op,qr;char ch;
    template<class I>
    il void gi(I &w)
    {
        ch=getchar(),op=1,w=0;
	    while(!isdigit(ch)){if(ch=='-') op=-1;ch=getchar();}
        while(isdigit(ch)){w=w*10+ch-'0';ch=getchar();}w*=op;
    }
    template<typename T,typename... Args>
    il void gi(T& t,Args&... args){gi(t);gi(args...);}
    template<class I>
    il void print(I w)
    {
        qr=0;if(!w) putchar('0');if(w<0) putchar('-'),w=-w;
	    while(w) que[++qr]=w%10+'0',w/=10;
        while(qr) putchar(que[qr--]);
    }
}
using io::gi;
using io::print;

const int N=2e7+5;
const int base=131;
const ll mod=19260817;

string a,b;
int tag[N];
ll ha[N],hb[N],fac[N];

ll gethash(int x)
{
    if(x+b.size()>a.size()) return -1;
    return (ha[x+b.size()-1]-ha[x-1]*fac[b.size()]%mod+mod)%mod;
}
int main()
{
    cin>>a>>b;
    while(a.size()<b.size()*2) a=a+a;
    a=a+a,a=a+a,a=a+a;
    fac[0]=1,ha[0]=a[0]-'a',hb[0]=b[0]-'a';
    for(int i=1;i<a.size();++i) fac[i]=fac[i-1]*base%mod;
    for(int i=1;i<a.size();++i) ha[i]=(ha[i-1]*base%mod+a[i]-'a')%mod;
    for(int i=1;i<b.size();++i) hb[i]=(hb[i-1]*base%mod+b[i]-'a')%mod;
    for(int i=0;i<a.size();++i) tag[i]=hb[b.size()-1]==gethash(i);
    int ans=0;
    for(int i=a.size()-b.size();i>=0;--i) if(tag[i]) tag[i]+=tag[i+b.size()],ans=max(ans,tag[i]);
    if(ans+1>=a.size()/b.size()) print(-1);
    else print(ans);
    end();
}