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: Accepted)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}
import scala.collection.mutable.{Map, Stack, Queue, PriorityQueue, Set}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val N, M = in.next().toInt
  val graph = Array.fill[List[Int]](N+1)(Nil)
  for (_ <- 1 to M) {
    val u, v = in.next().toInt
    graph(u) = v :: graph(u)
  }
  val S, T = in.next().toInt

  //
  val q = new Queue[(Int, Int)]() // (moveNum, node)
  val dp = Array.ofDim[Boolean](N+1, 4)

  q.enqueue((0, S))
  dp(S)(3) = true

  var ans = -1
  breakable {
    while (!q.isEmpty) {
      val (cost, now) = q.dequeue

      if (now == T) {
        ans = cost
        break
      }

      var set = Set[Int]()
      set += now

      for (i <- 0 to 3) {
        val nextSet = Set[Int]()
        for (e <- set) {
          if (i == 3) {
            if (!dp(e)(i)) {
              q.enqueue((cost+1, e))
            }
          }
          else {
            if (!dp(e)(i)) {
              dp(e)(i) = true
              for (e2 <- graph(e))
                nextSet += e2
            }
          }
        }

        set = nextSet
      }
    }
  }

  println(ans)

}

object Sub {
  def print2DimArray[A](mat: Array[Array[A]]): Unit = {
    mat.foreach(lin => println(lin.deep))
  }
}

class InputReader(stream: InputStream) {
  private[this] val reader = new BufferedReader(new InputStreamReader(stream))
  private[this] var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}