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: Accepted)

from collections import deque
import sys
input = sys.stdin.readline
def MI():return map(int,input().split())
def main():
  n,m=MI()
  G=[[] for _ in range(n)]
  for _ in range(m):
    u,v=MI()
    u-=1
    v-=1
    G[u].append(v)
  s,t=MI()
  s-=1
  t-=1
  
  fi=[True]*n
  se=[True]*n
  th=[True]*n
  th[s]=False
  
  dq=deque([])
  dq.append(s)
  depth=0
  
  while dq:
    depth+=1
    tank1=[]
    tank2=[]
    tank3=[]
    
    for p in dq:
      for c in G[p]:
        if fi[c]:
          fi[c]=False
          tank2.append(c)
    for p in tank2:
      for c in G[p]:
        if se[c]:
          se[c]=False
          tank3.append(c)
    for p in tank3:
      for c in G[p]:
        if th[c]:
          th[c]=False
          tank1.append(c)
        if c==t:
          print(depth)
          exit()
    dq=deque(tank1)
  print(-1)  
if __name__=='__main__':
  main()