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 defaultdict as dd
from collections import deque
N, M = map(int, input().split())
e = dd(list)
inf = 10 ** 9
for _ in range(M):
  u, v = map(int, input().split())
  e[u].append(v + N)
  e[u + N].append(v + 2 * N)
  e[u + 2 * N].append(v)
S, T = map(int, input().split())
S += 2 * N
T += 2 * N
stopped = [inf] * (3 * (N + 1))
Q = deque()
Q.append(S)
stopped[S] = 0
while len(Q) != 0:
  p = Q.popleft()
  if stopped[p] >= stopped[T]:
    break
  for q in e[p]:
    if stopped[q] > stopped[p] + 1:
      Q.append(q)
      stopped[q] = stopped[p] + 1
if stopped[T] == inf:
  print(-1)
else:
  print(stopped[T] // 3)