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 Scala (2.11.7) (judge verdict: Accepted)

object Main {
  import java.io.{BufferedReader, InputStream, InputStreamReader}
  import java.util.StringTokenizer
  import scala.reflect.ClassTag

  def main(args: Array[String]): Unit = {
    val out = new java.io.PrintWriter(System.out)
    new Main(out, new InputReader(System.in)).solve()
    out.flush()
  }

  private[this] val oj = System.getenv("ATCODER_DEBUG") == null
  @inline private final def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }
  @inline private final def debug(as: Array[Boolean]): Unit = if (!oj){ debug(as.map(x => if(x) "1" else "0").mkString) }
  @inline private final def debug(as: Array[Int]): Unit = if (!oj){ debug(as.mkString(" ")) }
  @inline private final def debug(as: Array[Long]): Unit =if (!oj){ debug(as.mkString(" ")) }
  @inline private final def debugDim(m: Array[Array[Long]]): Unit = if (!oj){
    REP(m.length) { i =>
      debug(m(i))
    }
  }
  @inline private final def debugDimFlip(m: Array[Array[Long]]): Unit = if (!oj){
    REP(m(0).length) { j =>
      REP(m.length) { i =>
        System.err.print(m(i)(j))
        System.err.print(" ")
      }
      System.err.println()
    }
  }
  @inline private final def debug(s: => String): Unit = DEBUG {
    System.err.println(s)
  }
  @inline private final def debugL(num: => Long): Unit = DEBUG {
    System.err.println(num)
  }
  private def isDebug[A](debug: => A, online: => A): A = {
    if (oj) online else debug
  }

  class InputReader(val stream: InputStream) {
    private[this] val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private[this] var tokenizer: StringTokenizer = _

    @inline private[this] final def next(): String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = Integer.parseInt(next())
    def nextLong(): Long = java.lang.Long.parseLong(next())
    def nextChar(): Char = next().charAt(0)

    def ni(): Int = nextInt()
    def nl(): Long = nextLong()
    def nc(): Char = nextChar()
    def ns(): String = next()
    def ns(n: Int): Array[Char] = ns().toCharArray
    def na(n: Int, offset: Int = 0): Array[Int] = map(n)(_ => ni() + offset)
    def na2(n: Int, offset: Int = 0): (Array[Int], Array[Int]) = {
      val A1, A2 = Array.ofDim[Int](n)
      REP(n) { i =>
        A1(i) = ni() + offset
        A2(i) = ni() + offset
      }
      (A1, A2)
    }
    def nm(n: Int, m: Int): Array[Array[Int]] = {
      val A = Array.ofDim[Int](n, m)
      REP(n) { i =>
        REP(m) { j =>
          A(i)(j) = ni()
        }
      }
      A
    }
    def nal(n: Int): Array[Long] = map(n)(_ => nl())
    def nm_c(n: Int, m: Int): Array[Array[Char]] = map(n) (_ => ns(m))
  }

  def REP(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = offset
    val N = n + offset
    while(i < N) { f(i); i += 1 }
  }
  def REP_r(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = n - 1 + offset
    while(i >= offset) { f(i); i -= 1 }
  }
  def TO(from: Int, to: Int)(f: Int => Unit): Unit = {
    REP(to - from + 1, from)(f)
  }
  def map[@specialized A: ClassTag](n: Int, offset: Int = 0)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    REP(n)(i => res(i) = f(i + offset))
    res
  }

  def sumL(as: Array[Int]): Long = {
    var s = 0L
    REP(as.length)(i => s += as(i))
    s
  }
  def cumSum(as: Array[Int]): Array[Long] = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}

object Workspace {
  import Main._
  class UnionFind(val n: Int) {
    private val par = Array.ofDim[Int](n)
    REP(n) (i => par(i) = i)
    val rank: Array[Int] = Array.fill[Int](n)(1) // 集合の要素数
    private val visits = Array.ofDim[Int](n) // 訪れた場所をfind毎に用意するのがもったいないのでつかいまわす

    def find(x: Int): Int = {
      var ptr = 0
      def step(x: Int): Int = {
        if (par(x) == x) x
        else {
          visits(ptr) = x
          ptr += 1
          step(par(x))
        }
      }

      val res = step(x)
      REP(ptr){ i => par(visits(i)) = res }
      res
    }

    private def merge(node: Int, rt: Int): Int = {
      par(node) = rt
      rank(rt) += rank(node)
      rt
    }

    def unite(x: Int, y: Int): Boolean = {
      val x1 = find(x)
      val y1 = find(y)
      if (x1 == y1) false
      else {
        if (rank(x1) < rank(y1))
          merge(x1, y1)
        else
          merge(y1, x1)

        true
      }
    }

    /**
      * xを解決する必要がないときは直にrankをみる
      */
    def cntNodes(x: Int): Int = rank(find(x))
  }
  class KMP(word: String) {
    val kmp: Array[Int] = Array.ofDim[Int](word.length + 1)
    2 to word.length foreach { i =>
      // kmp(i-1)以下しか参照されないからOK
      kmp(i) = longestPrefix(kmp(i - 1), word(i - 1))
    }

    def findFirst(text: String): Int = {
      var j = 0
      REP(text.length) { i =>
        j = longestPrefix(j, text(i))
        if (j == word.length) return i - word.length + 1
      }
      -1
    }

    def longestPrefix(matched: Int, c: Char): Int = {
      var j = matched
      var continues = true
      while(continues) {
        if (j < word.length && word(j) == c) {
          j += 1
          continues = false
        } else if (j == 0) continues = false
        else j = kmp(j)
      }
      j
    }
  }
}

class Main(out: java.io.PrintWriter, sc: Main.InputReader) {
  import sc._
  import Main._
  import java.util.Arrays.sort

  import scala.collection.mutable
  import math.{abs, max, min}
  import mutable.ArrayBuffer
  import Workspace._

  // toIntとか+7とかするならvalにしろ
  final private[this] val MOD = 1000000007

  def solve(): Unit = {
    val s, t = ns()
    val n = s.length
    val m = t.length
    val ss = new StringBuffer
    while(ss.length() < n + m) ss.append(s)
    val kmp = new KMP(t)
    var matched = 0
    val ok = Array.ofDim[Boolean](n)
    REP(ss.length) { i =>
      matched = kmp.longestPrefix(matched, ss.charAt(i))
      debug(s"$i $matched")
      if (matched == m) {
        ok((i - m + 1) % n) = true
      }
    }
    debug(ok)
    val uf = new UnionFind(n)
    REP(n) { i =>
      if (ok(i)) {
        if (!uf.unite(i, (i + m) % n)) {
          out.println(-1)
          return
        }
      }
    }
    var ans = 0
    val used = Array.ofDim[Boolean](n)
    REP(n) { i =>
      if (!used(uf.find(i))) {
        used(uf.find(i)) = true
        ans = max(ans, uf.cntNodes(i) - 1)
      }
    }
    out.println(ans)
  }
}