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.util
import java.util.Scanner

import scala.collection.mutable.ArrayBuffer


object Main {

  case class Edge(from: Int, to: Int)

  def read() = {
    val sc = new Scanner(System.in)
    val nVertex, nEdge = sc.nextInt()
    val edges = IndexedSeq.fill(nEdge) {
      val from = sc.nextInt() - 1
      val to = sc.nextInt() - 1
      Edge(from, to)
    }
    val start = sc.nextInt() - 1
    val goal = sc.nextInt() - 1
    (nVertex, nEdge, edges, start, goal)
  }

  def solve(nVertex: Int,
            nEdge: Int,
            edges: IndexedSeq[Edge],
            start: Int,
            goal: Int): Long = {
    val next1List = IndexedSeq.fill(nVertex)(ArrayBuffer.empty[Int])
    for (edge <- edges) {
      next1List(edge.from).append(edge.to)
    }
    val dist = IndexedSeq.fill(3)(Array.fill(nVertex)(Int.MaxValue / 2))
    val visited = IndexedSeq.fill(3)(Array.fill(nVertex)(false))
    dist(0)(start) = 0
    visited(0)(start) = true

    case class Info(v: Int, mod: Int)

    val queue: util.Queue[Info] = new util.ArrayDeque[Info]()
    queue.add(Info(start, 0))

    while (!queue.isEmpty) {
      val current = queue.remove()
      for (next <- next1List(current.v)) {
        val nextMod = (current.mod + 1) % 3
        if (!visited(nextMod)(next)) {
          dist(nextMod)(next) = dist(current.mod)(current.v) + (if (nextMod == 0) 1 else 0)
          visited(nextMod)(next) = true
          queue.add(Info(next, nextMod))
        }
      }
    }
    if (visited(0)(goal)) dist(0)(goal) else -1
  }

  def main(args: Array[String]): Unit = {
    val (nVertex, nEdge, edges, start, goal) = read()
    println(solve(nVertex, nEdge, edges, start, goal))
  }
}