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 Julia (0.5.0) (judge verdict: Time Limit Exceeded)

parseInt(x::SubString{String}) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function heap(a::Array{Tuple{Int,Int},1})
	h = [a[1]]
	for i in 2:length(a)
		hpush(h,a[i])
	end
	h
end

function hpush(h::Array{Tuple{Int,Int},1},x)
	push!(h,x)
	if length(h) > 1
		l = length(h)
		k = div(l,2)
		while h[k][2]>x[2]&&k>=1
			tmp = h[k]
			h[k] = x
			h[l] = tmp
			k = max(div(k,2),1)
			l = div(l,2)
		end
	end
	h
end

function hpop(h::Array{Tuple{Int,Int},1})
	x = h[1]
	z = pop!(h)
	if !isempty(h)
		h[1] = z
		k = 1
		l = length(h)
		while k <= l
			if 2*k>=l
				break
			elseif 2*k+1>=l
				if h[2*k][2]>=z[2]
					break
				else
					tmp = h[2*k]
					h[2*k] = z
					h[k] = tmp
					k = 2*k
				end
			else
				if h[2*k][2]>=z[2]&&h[2*k+1][2]>=z[2]
					break
				else
					t = h[2*k][2]<h[2*k+1][2]?2*k:2*k+1
					tmp = h[t]
					h[t] = z
					h[k] = tmp
					k = t
				end
			end
		end
	end
	x
end


function main()
	n,m = readline() |> split |> parseMap
	e = [Int[] for i in 1:3*n]
	for i in 1:m
		u,v = readline() |> split |> parseMap
		push!(e[u],n+v)
		push!(e[n+u],2*n+v)
		push!(e[2*n+u],v)
	end
	s,t = readline() |> split |> parseMap
	dis = -ones(Int,3*n)
	h = Array{Tuple{Int,Int},1}()
	push!(h,(s,0))
	while !isempty(h)
		v,w = hpop(h)
		dis[v] = w
		for i in 1:length(e[v])
			if dis[e[v][i]]<0 || dis[e[v][i]]>w+1
				hpush(h,(e[v][i],w+1))
			end
		end
	end
	println(dis[t]>=0?div(dis[t],3):-1)

end

main()