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 UnionFind

  attr_accessor :nodes, :sizes

  def initialize(num)
    self.nodes = []
    self.sizes = []

    num.times do |n|
      self.nodes[n] = n
      self.sizes[n] = 1
    end
  end

  def root(i)
    # Loop up the chain until reaching root
    while nodes[i] != i do
      # path compression for future lookups
      nodes[i] = nodes[nodes[i]]
      i = nodes[i]
    end
    i
  end

  def union(i, j)
    rooti = root i
    rootj = root j

    # already connected
    return if rooti == rootj

    # root smaller to root of larger
    if sizes[i] < sizes[j]
      nodes[rooti] = rootj
      sizes[rootj] += sizes[rooti]
    else
      nodes[rootj] = rooti
      sizes[rooti] += sizes[rootj]
    end
  end

  def connected?(i, j)
    root(i) == root(j)
  end

end

N, M = gets.split.map &:to_i

uf = UnionFind.new(N)

pathes = N.times.map{ [] }
M.times{
  from, to = gets.split.map(&:to_i).map &:pred
  pathes[from] << to
}
S, T = gets.split.map(&:to_i).map &:pred

# p pathes

kpathes = N.times.map{ [] }
N.times{|i|
  f = -> from, n {
    if n < 1
      kpathes[i] << from
      # kpathes[from] << i
      return
    end
    pathes[from].each{|to|
      f[to, n - 1]
    }
  }
  f[i, 3]
}
kpathes.map! &:uniq
# p kpathes

N.times{|from|
  kpathes[from].each{|to|
    uf.union(from, to)
  }
}
if !uf.connected?(S, T)
  p -1
  exit
end

ans = 0
q = {}
q[S] = 1
used = [false] * N
used[S] = true
while !q.empty?
  ans += 1
  new_q = {}
  q.each_key{|from|
    kpathes[from].each{|to|
      next if used[to]
      if to == T
        p ans
        exit
      end
      new_q[to] = 1
    }
  }
  q = new_q
end
p -1