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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class PriorityQueue

  def initialize(heap=[])
    @size = heap.size
    @heap = heap.sort
  end

  # log( log n )
  def push x
    i = @size
    @size += 1
    while i > 0
      par = ( i - 1 ) / 2
      break if @heap[par] <= x
      @heap[i] = @heap[par]
      i = par
    end
    @heap[i] = x
  end

  # log( log n )
  # return nil if @size == 0
  def pop

    ret = @heap[0]
    x = @heap[@size-=1]


    i = 0
    while (child = i * 2 + 1) < @size
      child_1 = i * 2 + 2
      child = child_1 if child_1 < @size && @heap[child_1] < @heap[child]
      break if @heap[child] >= x
      @heap[i] = @heap[child]
      i = child
    end

    @heap[i] = x
    @heap.delete_at(-1)
    ret
  end

  def size
    @heap.size
  end

  def top
    @heap[0]
  end

  def any?
    @size > 0
  end

  def empty?
    @size == 0
  end

  def all
    @heap
  end

  def first
    @heap[0]
  end

  def second
    @size < 3 ? @heap[1] : [@heap[1], @heap[2]].min
  end

  def third
    @size < 3 ? nil : [@heap[1], @heap[2]].max
  end
end

class Array

  include Comparable

  def to_pq
    PriorityQueue.new(self)
  end
end

class Node

  attr_accessor :to, :cost
  def initialize(to, cost)
    @to = to
    @cost = cost
  end
end

INF = Float::INFINITY
# with PriorityQueue O(|E|log|V|)
def dijkstra(g, s)

  # inf = 10 ** 12
  inf = INF
  dist = Array.new(g.size, inf)
  dist[s] = 0

  que = PriorityQueue.new
  que.push( [0, s] )

  while que.any?

    d, v = que.pop
    next if dist[v] < d
    # p g[v]
    g[v].each do |e|
      next if dist[v] == inf
      if dist[e.to] > dist[v] + e.cost
        dist[e.to] = dist[v] + e.cost
        que.push( [dist[e.to], e.to ] )
      end
    end
    # p que
  end

  dist
  # dist.map!{|d| d == inf ? "INF" : d }
end

mod = 10 ** 9 + 7

n, m = gets.to_s.split.map{|t|t.to_i}

g = Array.new(n*3){ [] }

m.times do
  u, v = gets.to_s.split.map{|t|t.to_i-1}
  g[3*u+0] << Node.new(3*v+1, 1)
  g[3*u+1] << Node.new(3*v+2, 1)
  g[3*u+2] << Node.new(3*v+0, 1)
  # g[3*v+0] << Node.new(3*u+1, 1)
  # g[3*v+1] << Node.new(3*u+2, 1)
  # g[3*v+2] << Node.new(3*u+0, 1)
end

s, t = gets.to_s.split.map{|i|i.to_i-1}

dist = dijkstra(g, s*3+0)
# p [s*3+0, t*3+0]

puts dist[t*3+0] == INF ? -1 : dist[t*3+0]/3