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;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

public class ABC
{
    //          long[] sp = Console.ReadLine().Split().Select(long .Parse).ToArray();
    //          int[] sp = Console.ReadLine().Split().Select(int.Parse).ToArray();
    public static void Main()
    {
        string s = Console.ReadLine();
        string t = Console.ReadLine();

        string ss = s + s;
        string sss = ss + s;
        ss=ss.Replace(t, "M");
        sss= sss.Replace(t, "M");

        long maxmaruss = 0;
        long tempmaxmaruss = 0;
        for (int i = 0; i < ss.Length; i++)
        {
            if (ss[i].ToString() == "M")
            {
                tempmaxmaruss = 0;
                for (; i < ss.Length; i++)
                {

                    if (ss[i].ToString() == "M")
                    {
                        tempmaxmaruss++;
                    }
                    else
                    {
                        if (tempmaxmaruss > maxmaruss)
                            maxmaruss = tempmaxmaruss;
                        tempmaxmaruss = 0;
                    }
                }
            }

        }

        long tempmaxmarusss = 0;
        long maxmarusss = 0;

        for (int i = 0; i < sss.Length; i++)
        {
            if (sss[i].ToString() == "M")
            {
                tempmaxmarusss = 0;
                for (; i < sss.Length; i++)
                {

                    if (sss[i].ToString() == "M")
                    {
                        tempmaxmarusss++;
                    }
                    else
                    {
                        if (tempmaxmarusss > maxmarusss)
                            maxmarusss = tempmaxmarusss;
                        tempmaxmarusss = 0;
                    }
                }
            }

        }

        if (maxmaruss == maxmarusss)
        {
            Console.WriteLine(maxmaruss);
        }
        else
        {
            Console.WriteLine(-1);
        }

    }


}








class Util
{
    public static long GCD(long a, long b)
    {
        if (a < b)
            swap(ref a, ref b);
        if (a % b == 0) return b;
        return GCD(b, a % b);

    }
    public static int GCD(int a, int b)
    {
        if (a < b)
            swap(ref a, ref b);
        if (a % b == 0) return b;
        return GCD(b, a % b);
    }

    public static void swap<T>(ref T a, ref T b)
    {
        T temp = b;
        b = a;
        a = temp;

    }
    public static long LCM(long a, long b)
    {
        return a * b / GCD(a, b);
    }
    public static long LCM(int a, int b)
    {
        return Math.BigMul(a, b) / GCD(a, b);
    }


    public static int M = (int)(Math.Pow(10,9))+7;
    private static int[] factorial_modM;
    public static int Multiple_modM(int a, int b)
    {
        return (int)(Math.BigMul(a, b) % M);
    }
    public static void factorial_modM_Prepare(int n)
    {
        factorial_modM = new int[n + 1];
        factorial_modM[0] = 1;
        for (int i = 1; i <= n; ++i)
        {
            factorial_modM[i] = Multiple_modM(factorial_modM[i - 1], i);
        }
    }
    public static int Factorial(int n)
    {
        return factorial_modM[n];
    }
    public static int Pow(int a, int m)
    {
        switch (m)
        {
            case 0:
                return 1;
            case 1:
                return a;
            default:
                int p1 = Pow(a, m / 2);
                int p2 = Multiple_modM(p1, p1);
                return ((m % 2) == 0) ? p2 : Multiple_modM(p2, a);
        }
    }
    public static int Div(int a, int b)
    {
        return Multiple_modM(a, Pow(b, M - 2));
    }
    public static int nCr_modM(int n, int r)
    {
        if (n < r) { return 0; }
        if (n == r) { return 1; }
        int res = Factorial(n);
        res = Div(res, Factorial(r));
        res = Div(res, Factorial(n - r));
        return res;
    }

}

class UnionFind<T>
{
    public int tree_height;
    public UnionFind<T> parent;
    public T item
    {
        get;
        private set;
    }

    public UnionFind(T item)
    {
        tree_height = 0;
        parent = this;
    }

    public UnionFind<T> FindAdam()
    {
        if (parent == this) return this;
        else
        {
            UnionFind<T> ParentOfParent = parent.FindAdam();
            parent = ParentOfParent;//縦長な構成をつなぎ直している
            return ParentOfParent;
        }
    }
    public static void Unite(UnionFind<T> a, UnionFind<T> b)
    {
        UnionFind<T> ParentOfA = a.FindAdam();
        UnionFind<T> ParentOfB = b.FindAdam();
        if (ParentOfA.tree_height < ParentOfB.tree_height)
        {
            ParentOfA.parent = ParentOfB;
            ParentOfB.tree_height = Math.Max(ParentOfA.tree_height + 1, ParentOfB.tree_height);
        }
        else
        {
            ParentOfB.parent = ParentOfA;
            ParentOfA.tree_height = Math.Max(ParentOfB.tree_height + 1, ParentOfA.tree_height);
        }

    }

}

static class Permutation<T>
{
    private static void Search(List<T[]> perms, Stack<T> stack, T[] a)
    {
        int N = a.Length;
        if (N == 0)
        {
            perms.Add(stack.Reverse().ToArray());
        }
        else
        {
            var b = new T[N - 1];
            Array.Copy(a, 1, b, 0, N - 1);
            for (int i = 0; i < a.Length; ++i)
            {
                stack.Push(a[i]);
                Search(perms, stack, b);
                if (i < b.Length) { b[i] = a[i]; }
                stack.Pop();
            }
        }
    }
    public static IEnumerable<T[]> All(IEnumerable<T> src)
    {
        var perms = new List<T[]>();
        Search(perms, new Stack<T>(), src.ToArray());
        return perms;
    }
}