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 'set'

str = gets.split
N,M = str[0].to_i, str[1].to_i

r = Array.new(N+1){Array.new(3){Set.new}}

for i in 1..M do
    str = gets.split
    u,v = str[0].to_i, str[1].to_i
    r[u][0].add(v)
end

str = gets.split
S,T = str[0].to_i, str[1].to_i

for i in 1..N do
    if r[i][0].size > 0
        r[i][0].each do |val|
            r[i][1].merge(r[val][0])
        end
    end
end
for i in 1..N do
    if r[i][1].size > 0
        r[i][1].each do |val|
            r[i][2].merge(r[val][0])
        end
    end
end

arrive = Set.new
narrive = Set.new
parrive = Set.new
empty = Set.new
arrive.add(S)
narrive.add(S)
answer = 1
while 1 > 0
    if narrive.size > 0
        narrive.each do |val|
            parrive.merge(r[val][2])
        end
    else
        puts -1
        exit
    end
    if parrive.include?(T)
        puts answer
        exit
    else
        narrive = parrive - arrive
        arrive.merge(parrive)
        parrive = parrive & empty
    end
    answer += 1
end