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: Runtime Error)

import sys
import itertools

N, M = map(int, input().split())
nexts = dict()
for _ in range(M):
    u, v = map(int, input().split())
    nexts.setdefault(u, [])
    nexts[u].append(v)
S, T = map(int, input().split())

if S not in nexts.keys():
    print(-1)
    sys.exit()
dist = 0
dists = [-1] * 3
queue = [S]
while len(queue) > 0:
    next_queue = []
    for n in queue:
        if n == T:
            m = dist % 3
            if m == 0:
                print(dist // 3)
                sys.exit()
            elif dists[m] == -1:
                dists[m] = dist
        else:
            next_queue.extend(nexts[n])
    if dists[1] > 0 and dists[2] > 0:
        break
    queue = next_queue
    dist += 1

if T not in nexts.keys():
    print(-1)
    sys.exit()
dist = 0
dists_rev = [-1] * 3
queue = [T]
while len(queue) > 0:
    next_queue = []
    for n in queue:
        if n == S:
            m = dist % 3
            if dists_rev[m] == -1:
                dists_rev[m] = dist
        else:
            next_queue.extend(nexts[n])
    if dists_rev[0] > 0 and dists_rev[1] > 0 and dists_rev[2] > 0:
        break
    queue = next_queue
    dist += 1

ans = -1
if dists[1] > 0 and dists_rev[1] > 0:
    a = (dists[1] * 2 + dists_rev[1]) // 3
    ans = max(ans, a)
if dists[2] > 0 and dists_rev[2] > 0:
    a = (dists[2] * 2 + dists_rev[2]) // 3
    ans = max(ans, a)
if dists[1] > 0 and dists_rev[0] > 0:
    a = (dists[1] * 3 + dists_rev[0] * 2) // 3
    ans = max(ans, a)
if dists[2] > 0 and dists_rev[0] > 0:
    a = (dists[1] * 3 + dists_rev[2] * 2) // 3
    ans = max(ans, a)
print(ans)