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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using SB = System.Text.StringBuilder;
//using System.Threading.Tasks;
//using System.Text.RegularExpressions;
//using System.Globalization;
//using System.Diagnostics;
using static System.Console;
using System.Numerics;
using static System.Math;
using pair = Pair<int, int>;

class Program
{
    static void Main()
    {
        //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
        new Program().solve();
        Out.Flush();
    }
    readonly Scanner cin = new Scanner();
    readonly int[] dd = { 0, 1, 0, -1, 0 }; //→↓←↑
    readonly int mod = 1000000007;
    readonly int dom = 998244353;
    bool chmax<T>(ref T a, T b) where T : IComparable<T> { if (a.CompareTo(b) < 0) { a = b; return true; } return false; }
    bool chmin<T>(ref T a, T b) where T : IComparable<T> { if (b.CompareTo(a) < 0) { a = b; return true; } return false; }

    void solve()
    {
        var S = ReadLine();
        var T = ReadLine();
        int N = S.Length;
        int M = T.Length;

        while (S.Length < N + M - 1) S += S;

        var hash = new RollingHash(T)[0, M];
        var A = new RollingHash(S);

        var ok = new bool[N];      

        for (int i = 0; i < N; i++)
        {
            ok[i] = A[i, i + M] == hash;
        }
        //ok.join();

        var U = new DisjointSet(N);
        long ans = 0;
        for (int i = 0; i < N; i++)
        {
            if (ok[i] && ok[(i + M) % N])
            {
                if (!U.Unite(i, (i + M) % N))
                {
                    WriteLine(-1);
                    return;
                }
                chmax(ref ans, U.Size(i));
            }
        }

        WriteLine(ans);
    }

}

class RollingHash
{
    const int BASE = 9973;
    readonly int[] mod = new[] { 999999937, 1000000007 };
    readonly long[][] hash, power;

    public RollingHash(string s)
    {
        hash = new long[2][];
        power = new long[2][];
        for (int iter = 0; iter < mod.Length; iter++)
        {
            hash[iter] = new long[s.Length + 1];
            power[iter] = new long[s.Length + 1];
            power[iter][0] = 1;
            for (int i = 0; i < s.Length; i++)
            {
                hash[iter][i + 1] = (hash[iter][i] * BASE + s[i]) % mod[iter];
                power[iter][i + 1] = power[iter][i] * BASE % mod[iter];
            }
        }
    }
    /// <summary>
    /// [l, r)のハッシュ値 0-indexed
    /// </summary>
    /// <param name="l"></param>
    /// <param name="r"></param>
    /// <param name="id"></param>
    /// <returns></returns>
    public long this[int l, int r, int id = 0]
    {
        get
        {
            long ret = hash[id][r] - hash[id][l] * power[id][r - l] % mod[id];
            if (ret < 0) ret += mod[id];
            return ret;
        }
    }
}

/// <summary>
/// 高速だが、悪意のあるケースでハッシュ衝突
/// </summary>
class RollingHash64
{
    const int BASE = 9973;
    readonly long[] hash, power;

    public RollingHash64(string s)
    {
        hash = new long[s.Length + 1];
        power = new long[s.Length + 1];
        power[0] = 1;
        for (int i = 0; i < s.Length; i++)
        {
            hash[i + 1] = hash[i] * BASE + s[i];
            power[i + 1] = power[i] * BASE;
        }
    }
    /// <summary>
    /// [l, r)のハッシュ値 0-indexed
    /// </summary>
    /// <param name="l"></param>
    /// <param name="r"></param>
    /// <returns></returns>
    public long this[int l, int r] => hash[r] - hash[l] * power[r - l];
}

/// <OriginalAuthor>camypaper</OriginalAuthor>
class DisjointSet
{
    readonly int[] par;
    readonly byte[] rank;
    /// <summary>
    /// 0 から <paramref name="N"/>-1 までの番号がついた <paramref name="N"/> 個の要素からなる集合を作成します．
    /// </summary>
    /// <param name="N">要素数</param>
    /// <remarks>このコンストラクタは O(N) で実行されます．</remarks>
    public DisjointSet(int N)
    {
        par = new int[N];
        rank = new byte[N];
        for (int i = 0; i < N; i++) par[i] = -1;
    }
    /// <summary>
    /// 指定した要素が属する集合の代表値を取得します．
    /// </summary>
    /// <param name="id">調べたい要素の 0-indexed での番号</param>
    /// <returns>指定した要素が属する集合の代表値</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public int this[int id]
    {
        get
        {
            if (par[id] < 0) return id;
            return par[id] = this[par[id]];
        }
    }
    /// <summary>
    /// 指定した 2 つの要素が属する集合同士を 1 つに統合することを試みます．
    /// </summary>
    /// <param name="x">最初の要素の 0-indexed での番号</param>
    /// <param name="y">2 つ目の要素の 0-indexed での番号</param>
    /// <returns>統合に成功したならば true，そうでなければ false．</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public bool Unite(int x, int y)
    {
        x = this[x]; y = this[y];
        if (x == y) return false;
        if (rank[x] < rank[y]) { var tmp = x; x = y; y = tmp; }
        par[x] += par[y];
        par[y] = x;
        if (rank[x] == rank[y])
            rank[x]++;
        return true;
    }
    /// <summary>
    /// 指定した要素が属する集合のサイズを取得します．
    /// </summary>
    /// <param name="x">指定する要素の 0-indexed での番号</param>
    /// <returns>集合のサイズ</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public int Size(int x) => -par[this[x]];

    /// <summary>
    /// 指定した 2 つの要素が属する集合が同じかどうかを判定する．計算量 O(α(N))．
    /// </summary>
    /// <param name="x">最初の要素の 0-indexed での番号</param>
    /// <param name="y">2 つ目の要素の 0-indexed での番号</param>
    /// <returns>同じならば true，そうでなければ false．</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public bool Same(int x, int y) => this[x] == this[y];
}

static class Ex
{
    public static void join<T>(this IEnumerable<T> values, string sep = " ") => WriteLine(string.Join(sep, values));
    public static string concat<T>(this IEnumerable<T> values) => string.Concat(values);
    public static string reverse(this string s) { var t = s.ToCharArray(); Array.Reverse(t); return t.concat(); }

    public static int lower_bound<T>(this IList<T> arr, T val) where T : IComparable<T>
    {
        int low = 0, high = arr.Count;
        int mid;
        while (low < high)
        {
            mid = ((high - low) >> 1) + low;
            if (arr[mid].CompareTo(val) < 0) low = mid + 1;
            else high = mid;
        }
        return low;
    }
    public static int upper_bound<T>(this IList<T> arr, T val) where T : IComparable<T>
    {
        int low = 0, high = arr.Count;
        int mid;
        while (low < high)
        {
            mid = ((high - low) >> 1) + low;
            if (arr[mid].CompareTo(val) <= 0) low = mid + 1;
            else high = mid;
        }
        return low;
    }
}

class Pair<T, U> : IComparable<Pair<T, U>> where T : IComparable<T> where U : IComparable<U>
{
    public T f; public U s;
    public Pair(T f, U s) { this.f = f; this.s = s; }
    public int CompareTo(Pair<T, U> a) => f.CompareTo(a.f) != 0 ? f.CompareTo(a.f) : s.CompareTo(a.s);
    public override string ToString() => $"{f} {s}";
}

class Scanner
{
    string[] s; int i;
    readonly char[] cs = new char[] { ' ' };
    public Scanner() { s = new string[0]; i = 0; }
    public string[] scan => ReadLine().Split();
    public int[] scanint => Array.ConvertAll(scan, int.Parse);
    public long[] scanlong => Array.ConvertAll(scan, long.Parse);
    public double[] scandouble => Array.ConvertAll(scan, double.Parse);
    public string next
    {
        get
        {
            if (i < s.Length) return s[i++];
            string st = ReadLine();
            while (st == "") st = ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
            return next;
        }
    }
    public int nextint => int.Parse(next);
    public long nextlong => long.Parse(next);
    public double nextdouble => double.Parse(next);
}