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 (7.3.0) (judge verdict: Wrong Answer)

import sys
input = sys.stdin.readline
from collections import deque


class Graph:
    def __init__(self, n, directed=False, decrement=True, destroy=False, edges=[]):
        self.n = n
        self.directed = directed
        self.decrement = decrement
        self.destroy = destroy
        self.edges = [set() for _ in range(self.n)]
        self.visited = [0]*self.n
        self.info = [-1]*self.n
        for x, y in edges:
            self.add_edge(x,y)

    def add_edge(self, x, y):
        if self.decrement:
            x -= 1
            y -= 1
        self.edges[x].add(y)
        if self.directed == False:
            self.edges[y].add(x)

    def add_adjacent_list(self, i, adjacent_list):
        if self.decrement:
            self.edges[i] = set(map(lambda x: x - 1, adjacent_list))
        else:
            self.edges[i] = set(adjacent_list)

    def bfs(self, start=1, goal=-1, time=0, save=False):
        """
        :param start: スタート地点
        :param goal: ゴール地点
        :param save: True = 前回の探索結果を保持する
        :return: （ループがあっても）最短距離。存在しなければ -1
        """
        if self.decrement:
            start -= 1
            goal -= 1
        if not save:
            self.visited = [0] * self.n
        p, t = start, time
        self.visited[p] = 1
        next_set = [(p, t)]

        while next_set:
            p, t = next_set.pop()
            for q in self.edges[p]:
                if (t+1)%3==0:
                    if self.visited[q] == 1:
                        continue
                    if q == goal:
                        return (t+1)//3
                    self.visited[q] = 1
                next_set.append((q, t + 1))
        return -1

##################################################################################################

N, M = map(int, input().split())
graph = Graph(N, directed=True, decrement=True, destroy=False)
for _ in range(M):
    x, y = map(int, input().split())
    graph.add_edge(x, y)
S, T = map(int, input().split())
print(graph.bfs(start=S, goal=T))