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.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2
N,M=map(int,input().split())
E=[[] for i in range(N)]
for i in range(M):
  u,v=map(int,input().split())
  u,v=u-1,v-1
  E[u].append(v)
S,T=map(int,input().split())
S,T=S-1,T-1

from collections import deque
stack=deque()
seen=[[False for i in range(3)] for j in range(N)]
stack.append([S,0,1])
while stack:
  node=stack.popleft()
  v=node[0]
  step=node[1]
  turn=node[2]
  if step==3:
    if v==T:
      print(turn)
      break
    step=0
    turn+=1
  if seen[v][step]:
    continue
  seen[v][step]=True
  children=E[v]
  for child in children:
    stack.append([child,step+1,turn])
else:
  print(-1)
