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: Runtime Error)

# require 'set'
N,M = gets.chomp.split(" ").map(&:to_i)
mat = []
M.times do
    mat << gets.chomp.split(" ").map(&:to_i)
end
S,T = gets.chomp.split(" ").map(&:to_i)

@graph = Array.new(N+1){[]}

# p [N,M]
# p mat
# p [S,T]

mat.each do |arr|
    u = arr[0]
    v = arr[1]
    @graph[u] << v
end
# p @graph
# (1..N).each do |n|

# next1 = []
=begin
@graph_2 = Array.new(N+1){Set.new}
(1..N).each do |n|
    @graph[n].each do |s1|
        @graph[s1].each do |s2|
            @graph[s2].each do |s3|
                if n != s3
                    @graph_2[n] << s3
                end
            end
        end
    end
end
=end
# p @graph_2
# next1.each do |n|
#     p n
# end
@visited = Array.new(N+1).map{Array.new(3)}
@q = []
@dist = Array.new(N+1).map{Array.new(3)}

def bfs(v)
    @visited[v][0] = true
    @dist[S][0]=0
    @q.push [v,0]

    while(tmp,s = @q.shift) do
        # break if tmp == T and @visited[n][dist_tmp % 3]
        # p [tmp,s]
        break if tmp == T and s == 0

        nodes = @graph[tmp]
        nodes.each do |n|
            dist_tmp = @dist[tmp][s]
            # p [tmp,dist_tmp]
            if not @visited[n][(s+1) % 3]
                @visited[n][(s+1) % 3] = true
                @q.push [n, (s+1) % 3]
                @dist[n][(s+1) % 3] = dist_tmp + 1
            end
        end
    end
end

bfs(S)
# p @dist
# p @dist
# p @visited
puts @dist[T][0] ? @dist[T][0] / 3 : -1