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.Globalization;
using System.IO;
using System.Linq;
using System.Text;

internal partial class Solver {

    private int GetMaxLoopString(string s, string t) {
        var isMatch = new bool[s.Length];
        foreach (var i in Kmp.GetMatchIndexes(s, t)) {
            isMatch[i] = true;
        }
        int result = 0;
        for (int i = 0; i < t.Length && i < s.Length; i++) {
            int count = 0;
            for (int j = i; j < s.Length; j += t.Length) {
                if (isMatch[j]) {
                    result = Math.Max(result, ++count);
                } else {
                    count = 0;
                }
            }
        }
        //new { s, t, result, isMatch }.Dump();
        return result;
    }

    public void Run() {
        var s = ns();
        var t = ns();

        var b = new StringBuilder();
        b.Append(s);
        while (b.Length <= t.Length) b.Append(s);
        s = b.ToString();

        var v1 = GetMaxLoopString(s + s, t);
        var v2 = GetMaxLoopString(s + s + s, t);
        int ans = 0;
        if (v1 < v2) ans = -1;
        else if (v1 == v2) ans = v1;
        else throw new Exception();
        cout.WriteLine(ans);
    }
}

public static class Kmp {
    public static int[] CreateKmpTable(string s) {
        int[] table = new int[s.Length];
        table[0] = 0;
        for (int i = 1; i < s.Length; i++) {
            int failure = table[i - 1];
            while (failure > 0 && s[i] != s[failure]) {
                failure = table[failure - 1];
            }

            if (s[i] == s[failure]) {
                failure++;
            }

            table[i] = failure;
        }
        return table;
    }

    public static List<int> GetMatchIndexes(string s, string pattern) {
        return GetMatchIndexes(s, CreateKmpTable(pattern), pattern);
    }

    public static List<int> GetMatchIndexes(string s, int[] kmpTable, string pattern) {
        int k = 0;
        var matchIndex = new List<int>();
        for (var i = 0; i < s.Length; i++) {
            while (k > 0 && pattern[k] != s[i]) {
                k = kmpTable[k - 1];
            }

            if (pattern[k] == s[i] && ++k == pattern.Length) {
                matchIndex.Add(i - pattern.Length + 1);
                k = kmpTable[k - 1];
            }
        }
        return matchIndex;
    }
}

// PREWRITEN CODE BEGINS FROM HERE
internal partial class Solver : Scanner {
    public static void Main(string[] args) {
#if LOCAL
        new Solver(Console.In, Console.Out).Run();
#else
        Console.SetOut(new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });
        new Solver(Console.In, Console.Out).Run();
        Console.Out.Flush();
#endif
    }

    public void RunMulti() {
        while (HasNext()) {
            Run();
        }
    }

#pragma warning disable IDE0052
    private readonly TextReader cin;
    private readonly TextWriter cout;
#pragma warning restore IDE0052

    public Solver(TextReader reader, TextWriter writer)
        : base(reader) {
        cin = reader;
        cout = writer;
    }
    public Solver(string input, TextWriter writer)
        : this(new StringReader(input), writer) {
    }

#pragma warning disable IDE1006
#pragma warning disable IDE0051
    private int ni() { return NextInt(); }
    private int[] ni(int n) { return NextIntArray(n); }
    private long nl() { return NextLong(); }
    private long[] nl(int n) { return NextLongArray(n); }
    private double nd() { return NextDouble(); }
    private double[] nd(int n) { return NextDoubleArray(n); }
    private string ns() { return Next(); }
    private string[] ns(int n) { return NextArray(n); }
#pragma warning restore IDE1006
#pragma warning restore IDE0051
}

public static class LinqPadExtension {
    public static T Dump<T>(this T obj) {
#if LOCAL
        return LINQPad.Extensions.Dump(obj);
#else
        return obj;
#endif
    }
}
public class Scanner {
    private readonly TextReader Reader;
    private readonly Queue<string> TokenQueue = new Queue<string>();
    private readonly CultureInfo ci = CultureInfo.InvariantCulture;

    public Scanner()
        : this(Console.In) {
    }

    public Scanner(TextReader reader) {
        Reader = reader;
    }

    public int NextInt() { return int.Parse(Next(), ci); }
    public long NextLong() { return long.Parse(Next(), ci); }
    public double NextDouble() { return double.Parse(Next(), ci); }
    public string[] NextArray(int size) {
        var array = new string[size];
        for (var i = 0; i < size; i++) {
            array[i] = Next();
        }

        return array;
    }
    public int[] NextIntArray(int size) {
        var array = new int[size];
        for (var i = 0; i < size; i++) {
            array[i] = NextInt();
        }

        return array;
    }

    public long[] NextLongArray(int size) {
        var array = new long[size];
        for (var i = 0; i < size; i++) {
            array[i] = NextLong();
        }

        return array;
    }

    public double[] NextDoubleArray(int size) {
        var array = new double[size];
        for (var i = 0; i < size; i++) {
            array[i] = NextDouble();
        }

        return array;
    }

    public string Next() {
        if (TokenQueue.Count == 0) {
            if (!StockTokens()) {
                throw new InvalidOperationException();
            }
        }
        return TokenQueue.Dequeue();
    }

    public bool HasNext() {
        if (TokenQueue.Count > 0) {
            return true;
        }

        return StockTokens();
    }

    private static readonly char[] _separator = new[] { ' ', '\t' };
    private bool StockTokens() {
        while (true) {
            var line = Reader.ReadLine();
            if (line == null) {
                return false;
            }

            var tokens = line.Split(_separator, StringSplitOptions.RemoveEmptyEntries);
            if (tokens.Length == 0) {
                continue;
            }

            foreach (var token in tokens) {
                TokenQueue.Enqueue(token);
            }

            return true;
        }
    }
}