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
def i1():
 return int(input())
def i2():
 return [int(i) for i in input().split()]
e=[[] for i in range(3*10**5+10)]

[n,m]=i2()
for i in range(m):
 [u,v]=i2()
 u-=1
 v-=1
 e[3*u].append(3*v+1)
 e[3*u+1].append(3*v+2)
 e[3*u+2].append(3*v)
 
[s,t]=i2()
s-=1
t-=1
from collections import deque
d=[-1 for i in range(3*10**5+10)]
def bfs(v):
   d[v]=0
   q=deque([v])
   while len(q):
    v=q.popleft()
    for i in e[v]:
      if d[i]<0:
         d[i]=d[v]+1
         q.append(i)
   if d[3*t]>-1:
     return d[3*t]//3
   return d[3*t]  
print(bfs(3*s))