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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using static System.Console;
using static System.Math;

namespace AtTest.ABC_135
{
    class F
    {
        static void Main(string[] args)
        {
            var sw = new System.IO.StreamWriter(OpenStandardOutput()) { AutoFlush = false };
            SetOut(sw);

            Method(args);

            Out.Flush();
        }

        static void Method(string[] args)
        {
            string s = Read();
            string t = Read();
            string ss = s;

            while (ss.Length < s.Length + t.Length - 1) ss += s;
            RollingHash rollingHash = new RollingHash(ss);
            long[] hash = rollingHash.GetStringHash(t);
            List<int>[] graph = new List<int>[s.Length];
            for (int i = 0; i < s.Length; i++) graph[i] = new List<int>();
            for (int i = 0; i < s.Length; i++)
            {
                if (rollingHash.ValidateRange(i, t.Length, hash))
                {
                    graph[i].Add((i + t.Length) % s.Length);
                }
            }
            int[] lengthes = new int[s.Length];
            for (int i = 0; i < s.Length; i++) lengthes[i] = -1;
            for(int i = 0; i < s.Length; i++)// from leaf
            {
                if (graph[i].Count > 0) continue;

                int cnt = 0;
                lengthes[i] = cnt;
                int now = (i - t.Length % s.Length + s.Length) % s.Length;
                while (graph[now].Count >0)
                {
                    cnt++;
                    lengthes[now] = cnt;
                    now= (now - t.Length % s.Length + s.Length) % s.Length;
                }
            }

            int res = 0;
            for(int i = 0; i < s.Length; i++)
            {
                if (lengthes[i] == -1)
                {
                    WriteLine(-1);
                    return;
                }

                res = Max(res, lengthes[i]);
            }
            WriteLine(res);
        }

        static int DFS(ref int[] lengthes,int now,List<int>[] graph)
        {
            if (lengthes[now] >= -1) return lengthes[now];

            if (graph[now].Count > 0)
            {
                int val = DFS(ref lengthes, graph[now][0], graph);
                if (val == -1) lengthes[now] = -1;
                else if (lengthes[now] >= 0) lengthes[now] = -1;
                else lengthes[now] = val + 1;
            }
            else
            {
                lengthes[now] = 0;
            }
            return lengthes[now];
        }

        class RollingHash
        {
            long mod = 1000000007;
            long[] bases = new long[3] { 1009, 2003, 3001 };
            long[,] pows;
            long[,] hashes;

            public RollingHash(string s)
            {
                pows = new long[s.Length, bases.Length];
                hashes = new long[s.Length, bases.Length];

                for (int i = 0; i < s.Length; i++)
                {
                    for (int j = 0; j < bases.Length; j++)
                    {
                        if (i == 0) pows[i, j] = 1;
                        else pows[i, j] = (pows[i - 1, j] * bases[j]) % mod;
                    }
                }

                for (int i = 0; i < s.Length; i++)
                {
                    for (int j = 0; j < bases.Length; j++)
                    {
                        if (i > 0) hashes[i, j] = (hashes[i - 1, j] * bases[j]) % mod;
                        hashes[i, j] += s[i];
                        hashes[i, j] %= mod;
                    }
                }
            }

            public long[] GetRangeHash(int l, int r)
            {
                long[] hash = new long[bases.Length];
                for (int i = 0; i < bases.Length; i++)
                {
                    hash[i] = hashes[r, i];
                    if (l > 0)
                    {
                        hash[i] += mod - (hashes[l - 1, i] * pows[r - l + 1, i]) % mod;
                        hash[i] %= mod;
                    }
                }
                return hash;
            }

            public bool ValidateRange(int l, int length, long[] hash)
            {
                int r = l + length - 1;
                long[] rangeHash = GetRangeHash(l, l + length - 1);
                for (int i = 0; i < bases.Length; i++)
                {
                    if (rangeHash[i] != hash[i]) return false;
                }
                return true;
            }

            public long[] GetStringHash(string s)
            {
                long[] vals = new long[bases.Length];
                for (int i = 0; i < s.Length; i++)
                {
                    for (int j = 0; j < bases.Length; j++)
                    {
                        vals[j] = (vals[j] * bases[j] + s[i]) % mod;
                    }
                }

                return vals;
            }

            public string ToHashString(long[] hash)
            {
                string val = "";
                for (int i = 0; i < bases.Length; i++) val += hash[i].ToString();
                return val;
            }
        }

        private static string Read() { return ReadLine(); }
        private static char[] ReadChars() { return Array.ConvertAll(Read().Split(), a => a[0]); }
        private static int ReadInt() { return int.Parse(Read()); }
        private static long ReadLong() { return long.Parse(Read()); }
        private static double ReadDouble() { return double.Parse(Read()); }
        private static int[] ReadInts() { return Array.ConvertAll(Read().Split(), int.Parse); }
        private static long[] ReadLongs() { return Array.ConvertAll(Read().Split(), long.Parse); }
        private static double[] ReadDoubles() { return Array.ConvertAll(Read().Split(), double.Parse); }
    }
}