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.chomp.split.map{|x| x.to_i}

nodes = {}

m.times {
	x, y = gets.chomp.split.map{|x| x.to_i}
	if nodes[x] == nil then
		nodes[x] = [y]
	else
		nodes[x] << y
	end
}

from, to = gets.chomp.split.map{|x| x.to_i}

zeros = {from => 0}
ones = {}
twoes = {}

is_detected = false

tmp_zeros = {from => 0}
tmp_ones = {}
tmp_twoes = {}

while tmp_zeros.keys.size + tmp_ones.keys.size + tmp_twoes.keys.size > 0 do
	tmp_tmp_zeros = {}
	tmp_tmp_ones = {}
	tmp_tmp_twoes = {}
	

	tmp_zeros.keys.each{ |k|
		nodes[k] && nodes[k].each{ |x|
			if ones[x] == nil then
				ones[x] = tmp_zeros[k]
				tmp_tmp_ones[x] = tmp_zeros[k]
			end
		}
	}

	tmp_ones.keys.each{ |k|
		nodes[k] && nodes[k].each{ |x|
			if twoes[x] == nil then
				twoes[x] = tmp_ones[k]
				tmp_tmp_twoes[x] = tmp_ones[k]
			end
		}
	}

	tmp_twoes.keys.each{ |k|
		nodes[k] && nodes[k].each{ |x|
			if zeros[x] == nil then
				zeros[x] = tmp_twoes[k] + 1
				tmp_tmp_zeros[x] = tmp_twoes[k] + 1
			end
		}
	}

	tmp_zeros = tmp_tmp_zeros
	tmp_ones = tmp_tmp_ones
	tmp_twoes = tmp_tmp_twoes

	if zeros[to] then
		puts zeros[to]
		is_detected = true
		break
	end
end

if ! is_detected then
	puts '-1'
end

STDOUT.flush