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
from collections import deque
n,m=map(int,input().split())
if m==0:
  print(-1)
  exit()
edge=[[] for i in range(3*n)]
for i in range(m):
  u,v=map(int,input().split())
  edge[3*(u-1)].append(3*(v-1)+1)
  edge[3*(u-1)+1].append(3*(v-1)+2)
  edge[3*(u-1)+2].append(3*(v-1))
s,t=map(int,input().split())
s-=1
t-=1
visited=[0 for i in range(3*m)]
def bfs(edge,visited,s,t):
  q=deque([3*s])
  visited[3*s]=1
  cnt=0
  while q:
    x=q.popleft()
    if x%3==0 and x//3==t:
      return visited[x]//3
    for i in edge[x]:
      newx=i
      if visited[newx]==0:
        visited[newx]=visited[x]+1
        q.append(newx)
  return -1
print(bfs(edge,visited,s,t))