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: Accepted)

N, M = gets.split.map(&:to_i)
edges = Array.new(N) { [] }

M.times do
  a, b = gets.split.map(&:to_i)
  # 頂点は全て処理しやすいように -1 する 
  edges[a - 1].push(b - 1)
end
# スタートとゴールも -1 する
S, T = gets.split.map(&:to_i).map(&:pred)

# 距離を格納するための配列
# 3の倍数、余り1、余り2のステップで到達した時の、最短距離をそれぞれ格納
dist = Array.new(N) { Array.new(3) }
que = [[S, 0]]
# BFSでdistを埋めていく
while !que.empty?
  start, step = que.shift
  # 3つのうちどこに格納すべきか判断するための変数
  next_step = (step + 1) % 3
  edges[start].each do |edge|
    # 任意の頂点、ステップ（3の倍数か、余り1か、余り2か）が埋まっていない場合のみ埋める
    # BFSで埋めているので、先に埋めた方が最短距離であるため更新はしない
    unless dist[edge][next_step]
      dist[edge][next_step] = step + 1
      # キューにスタート地点と今までにかかったステップ数を入れる
      que.push([edge, step + 1])
    end
  end
end

ans = dist[T][0] || -3
puts ans/3