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 (3.8.2) (judge verdict: Accepted)

from collections import defaultdict, deque


def kenkenpa(s, visited):
    res = []
    for v1 in e[s]:
        if visited[v1][1]:
            continue
        visited[v1][1] = True
        for v2 in e[v1]:
            if visited[v2][2]:
                continue
            visited[v2][2] = True
            for v3 in e[v2]:
                if visited[v3][0]:
                    continue
                visited[v3][0] = True
                res.append(v3)
    return res


N, M = map(int, input().split())
e = defaultdict(list)
for i in range(M):
    u, v = map(int, input().split())
    e[u].append(v)

S, T = map(int, input().split())
q = deque([[S, 0]])
visited = [[False]*3 for _ in range(N+1)]
visited[S][0] = True
ans = -1
while q:
    u, c = q.popleft()
    if u == T:
        ans = c
        break
    l = kenkenpa(u, visited)
    for v in l:
        q.append([v, c+1])
print(ans)