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)

# frozen_string_literal: true

n, m = gets.split.map(&:to_i)
uv = m.times.map { gets.split.map(&:to_i) }
s, t = gets.split.map(&:to_i)

d = Array.new(n + 1) { -1 }

ts = Array.new(n + 1) { [] }

uv.each do |e|
  u = e[0]
  v = e[1]
  ts[u].push v
end

$kh = {}

def kenken(s, ts, d, count, dest)
  return if $kh[s]

  ts[s].each do |t1|
    ts[t1].each do |t2|
      ts[t2].each do |t3|
        if d[t3].negative?
          d[t3] = count
          dest.push t3
        end
      end
    end
  end
  $kh[s] = true
  dest
end

count = 1
dest = [s]
loop do
  break if dest.empty?

  ndest = []
  dest.each do |ss|
    kenken(ss, ts, d, count, ndest)
  end
  if d[t].positive?
    puts d[t]
    exit
  end
  count += 1
  dest = ndest
end
puts '-1'