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
import queue

N, M = list(map(int, input().split()))
uv = [list(map(int, input().split())) for _ in range(M)]
S, T = list(map(int, input().split()))

V = [[0, 0, 0] for _ in range(N + 1)]
R = {}

for u, v in uv:
  if u not in R:
    R[u] = [v]
  else:
    R[u].append(v)
q = queue.Queue()
put = q.put
get = q.get

put([S, 0, 2])
ans = -1
while not q.empty():
  t = get()
  if t[0] == T and t[2] == 2:
    ans = t[1]
    break
  k = (t[2] + 1) % 3
  m = t[1]
  if t[2] == 2:
    m += 1
  if t[0] not in R: continue
  for i in R[t[0]]:
    if V[i][k] == 0:
      V[i][k] = 1
      put([i, m, k])

print(ans)