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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import heapq
N,M = map(int,input().split())
edge = [[] for i in range(N)]
double_edge = [[] for i in range(N)]
triple_edge = [[] for i in range(N)]

for i in range(M):
	u,v = map(int,input().split())
	u -= 1
	v -= 1
	edge[u].append(v)
S,T = map(int,input().split())
S -= 1
T -= 1
for i in range(N):
	e = edge[i]
	l = []
	for x in e:
		l += edge[x]
	double_edge[i] = list(set(l))
for i in range(N):
	e = edge[i]
	l = []
	for x in e:
		l += double_edge[x]
	l = list(set(l))
	if i in l:
		l.remove(i)
	triple_edge[i] = l

INF = 10**10
dist = [INF if i != S else 0 for i in range(N)]
search = []
searched = [False for i in range(N)]
heapq.heapify(search)
heapq.heappush(search, (0,S))
while len(search) > 0 and dist[T] == INF:
	x = heapq.heappop(search)
	d, node = x[0], x[1]
	if searched[node]:
		continue
	searched[S] = True
	dist[node] = d
	for i in triple_edge[node]:
		if not searched[i]:
			heapq.heappush(search, (min(d+1, dist[i]), i))
if dist[T] < INF:
	print(dist[T])
else:
	print(-1)