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: Wrong Answer)

# -*- coding: utf-8 -*-

def kkp_iter(s, neighbor, d, cnt):
    if cnt == 3:
        d.add(s)
        return

    for n in neighbor[s]:
        kkp_iter(n, neighbor, d, cnt+1)


def kkp(s, neighbor):
    d = set ()
    kkp_iter(s, neighbor, d, 0)
    return d


n, m = map(int, input().split())

neighbor = [[] for _ in range(n)]
for _ in range(m):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    neighbor[u].append(v)

s, t = map(int, input().split())
s -= 1
t -= 1

gone = set()
q = []

dest_t = False
kkp_cnt = 0
while not dest_t:
    gone.add(s)
    next_s = kkp(s, neighbor)
    kkp_cnt += 1

    for ns in next_s:
        if ns == t:
            dest_t = True
            break
        elif ns in gone:
            continue
        else:
            q.append(ns)

    if len(q) > 0:
        s = q.pop(0)
    else:
        break

if dest_t:
    print(kkp_cnt)
else:
    print(-1)