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())

g = [[] for i in range(n)]

for i in range(m):
  u, v = map(int, input().split())
  u, v = u-1, v-1
  g[u].append(v)
# print(g)

s, t = map(int, input().split())
s, t = s-1, t-1

visited = [[-1]*3 for j in range(n)]

l = 0
queue = deque([[s, l]])
visited[s][l] = 0
while queue:
  x, l = queue.popleft()
  if [x, l] == [t, 0]:
    print(visited[x][l]//3)
    exit()
  for x_ in g[x]:
    new_x, new_l = x_, (l+1)%3
    if visited[new_x][new_l] == -1:
      visited[new_x][new_l] = visited[x][l]+1
      queue.append([new_x, new_l])
    
print(visited[t][0])