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: Wrong Answer)

#!/usr/bin/python3
# -*- coding: utf-8 -*-
from collections import defaultdict



class Graph:

  def __init__(self,N,uv,S,T):
    self.graph = uv
    self.start = S
    self.goal = T
    self.states  = [[-1 for _ in range(3)] for _ in range(N)]
    self.states[self.start][0] = 0

  def solve(self):
    nxt = {self.start}
    visited = set([])
    while len(nxt) > 0 and len(visited) < len(self.states):
      p = nxt.pop()
      ## return
      if self.states[self.goal][0] >= 0:
        return self.states[self.goal][0]

      ## update
      for q in self.graph[p]:
        for ps in range(3):
          if self.states[p][ps] >= 0:
            qs = (ps+1)%3
            if self.states[q][qs] < 0:
              self.states[q][qs] = self.states[p][ps]+1
              nxt.add(q)
            else:
              self.states[q][qs] = min(self.states[q][qs],self.states[p][ps]+1)
    return -3


def main():
  ## input
  N,M = map(int, input().split())
  uv = defaultdict(set)
  for _ in range(M):
    u,v = map(int, input().split())
    uv[u-1].add(v-1)
  S,T = map(int, input().split())

  ## solve
  graph = Graph(N,uv,S-1,T-1)
  print(graph.solve()//3)


if __name__ == "__main__":
  main()