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: Runtime Error)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Diagnostics;
using Debug = System.Diagnostics.Trace;
using SB = System.Text.StringBuilder;
using System.Numerics;
using static System.Math;
using Number = System.Int32;
using CompLib.Strings;
namespace Program {
	public class Solver {
		Random rnd = new Random();
		public void Solve() {
			var s = rs;
			var t = rs;
			Swap(ref s, ref t);
			var k = (s.Length + t.Length - 1) / t.Length;
			var T = new char[t.Length * k];
			for (int i = 0; i < T.Length; i++)
				T[i] = t[i % t.Length];
			var pat = s + "$" + T.AsString() + T.AsString();
			Debug.WriteLine(s);
			Debug.WriteLine(T.AsString());
			Debug.WriteLine(pat.AsString());
			var Z = stringEx.ZAlgorithm(pat);
			var G = Enumerate(T.Length, x => -1);
			for (int i = 0; i < T.Length; i++) {
				var len = Z[i + s.Length + 1];
				if (len == s.Length) G[i] = (i + len) % T.Length;
			}
			Debug.WriteLine(G.AsJoinedString());
			var dp = Enumerate(T.Length, x => -1);
			var vis = new bool[T.Length];
			Func<int, int> dfs = null;
			dfs = cur => {
				if (dp[cur] != -1) return dp[cur];
				Debug.WriteLine(cur);
				if (vis[cur]) {
					Console.WriteLine(-1);
					Environment.Exit(0);
				}
				vis[cur] = true;
				var ret = 0;
				if (G[cur] != -1) ret = 1 + dfs(G[cur]);
				vis[cur] = false;
				return dp[cur] = ret;
			};
			var max = 0;
			for (int i = 0; i < T.Length; i++) max = Math.Max(max, dfs(i));
			Console.WriteLine(max);
		}


		const long INF = 1L << 60;
		static int[] dx = { -1, 0, 1, 0 };
		static int[] dy = { 0, 1, 0, -1 };
		int ri { get { return sc.Integer(); } }
		long rl { get { return sc.Long(); } }
		double rd { get { return sc.Double(); } }
		string rs { get { return sc.Scan(); } }
		public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

		static T[] Enumerate<T>(int n, Func<int, T> f) {
			var a = new T[n];
			for (int i = 0; i < a.Length; ++i) a[i] = f(i);
			return a;
		}
		static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
	}
}

#region main
static class Ex {
	static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
	static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
		return string.Join(st, ie);
	}
	static public void Main() {
		Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = false });
		var solver = new Program.Solver();
		var t = new System.Threading.Thread(solver.Solve, 10000000);
		t.Start();
		t.Join();
		//solver.Solve();
		Console.Out.Flush();
	}
}
#endregion
#region Ex
namespace Program.IO {
	using System.IO;
	using System.Text;
	using System.Globalization;

	public class Printer : StreamWriter {
		public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
		public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
	}

	public class StreamScanner {
		public StreamScanner(Stream stream) { str = stream; }

		public readonly Stream str;
		private readonly byte[] buf = new byte[1024];
		private int len, ptr;
		public bool isEof = false;
		public bool IsEndOfStream { get { return isEof; } }

		private byte read() {
			if (isEof) return 0;
			if (ptr >= len) {
				ptr = 0;
				if ((len = str.Read(buf, 0, 1024)) <= 0) {
					isEof = true;
					return 0;
				}
			}
			return buf[ptr++];
		}

		public char Char() {
			byte b = 0;
			do b = read(); while ((b < 33 || 126 < b) && !isEof);
			return (char)b;
		}
		public string Scan() {
			var sb = new StringBuilder();
			for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
			return sb.ToString();
		}
		public string ScanLine() {
			var sb = new StringBuilder();
			for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
			return sb.ToString();
		}
		public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
		public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
		public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
	}
}

#endregion


namespace CompLib.Strings {
	#region Z algorithm
	static public partial class stringEx {
		/// <summary> <paramref name="s"/> と <paramref name="s"/>[i:] の最長共通接頭辞を O(|<paramref name="s"/>|) で求める。</summary>
		static public int[] ZAlgorithm(string s) {
			var a = new int[s.Length + 1];
			a[0] = s.Length;
			int i = 1, j = 0;
			while (i < s.Length) {
				while (i + j < s.Length && s[j] == s[i + j]) ++j;
				a[i] = j;
				if (j == 0) { ++i; continue; }
				int k = 1;
				while (i + k < s.Length && k + a[k] < j) { a[i + k] = a[k]; ++k; }
				i += k; j -= k;
			}
			return a;
		}
	}
	#endregion
}