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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from itertools import chain
from collections import deque


def follow(edges, v, d):
    if d == 3:
        return [v]
    else:
        ans = [follow(edges, u, d + 1) for u in edges[v]]
        return list(chain.from_iterable(ans))


def main():
    N, M = map(int, input().split())
    edges = [[] for _ in range(N)]

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

    converted_edges = [[] for _ in range(N)]
    for v, v_edge in enumerate(edges):
        converted_edges[v] = follow(edges, v, 0)

    S, T = map(lambda x: int(x) - 1, input().split())
    q = deque([(S, 0)])

    nodes = [-1] * N

    while q:
        cv, count = q.popleft()
        if cv == T:
            if nodes[cv] == -1:
                nodes[cv] = count
            else:
                nodes[cv] = min(nodes[cv], count)
        else:
            if nodes[cv] == -1 or count < nodes[cv]:
                nodes[cv] = count
                for u in converted_edges[cv]:
                    q.append((u, count + 1))

    print(nodes[T])


if __name__ == '__main__':
    main()