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)

require 'pp'

N,M = readline.chomp.split.map(&:to_i)
g1 = Array.new(N) { [] }
M.times do |i|
  u,v = readline.chomp.split.map(&:to_i).map {|x| x-1}
  g1[u] << v
end
#pp g1

def dfs(g, i, d, ans)
  if d == 0
    ans << i
    return
  end

  g[i].each do |j|
    dfs(g, j, d-1, ans)
  end
end
$memo = {}
def neighbors(g,i)
  if $memo.has_key?(i)
    return $memo[i]
  end
  a = []
  dfs(g,i,3,a)
  $memo[i] = a
  return a
end

S,T = readline.chomp.split.map(&:to_i).map {|x| x-1}

cnt = 0
visited = {}
visited[S] = true
queue = []
queue.push(S)
d = Array.new(N,-1)
d[S] = 0
while v = queue.shift
  #pp v,queue
  if v == T
    break
  end
  neighbors(g1,v).each do |j|
    if not visited[j]
      visited[j] = true
      d[j] = d[v]+1
      queue.push(j)
    end
  end
end
puts d[T]