Problem:
Score : 500 points

Problem Statement

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G.
G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq \min(10^5, N (N-1))

1 \leq u_i, v_i \leq N(1 \leq i \leq M)

u_i \neq v_i (1 \leq i \leq M)

If i \neq j, (u_i, v_i) \neq (u_j, v_j).

1 \leq S, T \leq N

S \neq T

Input

Input is given from Standard Input in the following format:

N M
u_1 v_1
:
u_M v_M
S T

Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1.
If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2

Solution in Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import scala.io.StdIn._
import scala.collection.mutable.{ArrayBuffer, HashSet}
object Main extends App {
  val in = new StdinReader
  val n, m = in.nextInt
  val vs = Array.fill(n + 1)(new ArrayBuffer[Int]())
  (1 to m).foreach { i =>
    val u, v = in.nextInt
    vs(u) += v
  }
  val s, t = in.nextInt

  val arrived = new HashSet[Int]()

  def loop(cnt: Int, curs: Seq[Int]): Int = {
    val nexts = for {
      start <- curs
      hop <- vs(start)
      step <- vs(hop)
      jump <- vs(step)
      if !arrived.contains(jump)
    } yield {
      arrived += jump
      jump
    }

    if (nexts.contains(t)) cnt
    else if (nexts.isEmpty) -1
    else loop(cnt + 1, nexts)
  }

  println(loop(1, Seq(s)))
}


class StdinReader {
  val is = System.in
  val buffer = new Array[Byte](1024 * 1024)
  var size = 0
  var pos = 0

  private def readByte: Byte = {
    if (pos >= size) {
      pos = 0;
      size = is.read(buffer);
    }
    if (size < 0) -1
    else {
      val ret = buffer(pos)
      pos += 1
      ret
    }
  }

  // support single byte char only
  def next: String = {
    @scala.annotation.tailrec
    def r(sb: StringBuilder, reading: Boolean): String = {
      readByte match {
        case b if ('!' <= b && b <= '~') => r(sb.append(b.asInstanceOf[Char]), true)
        case _ => if (reading) sb.toString else r(sb, false)
      }
    }
    r(new StringBuilder, false)
  }

  def nextInt: Int = {
    @scala.annotation.tailrec
    def r(i: Int, reading: Boolean): Int = {
      readByte match {
        case '-' => r(i * -1, true)
        case b if '0' <= b && b <= '9' => r(i * 10 + (b - '0'), true)
        case _ => if (reading) i else r(i, false)
      }
    }
    r(0, false)
  }
}