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 main():
  N, M = map(int, input().split())
  edge = [[[] for _ in range(3)] for _ in range(N)]
  for i in range(M):
    u, v = map(int, input().split())
    edge[u-1][0].append([v-1,1])
    edge[u-1][1].append([v-1,2])
    edge[u-1][2].append([v-1,0])
  d = [[-1]*3 for _ in range(N)]
  S, T = map(int, input().split())
  d[S-1][0] = 0
  que = [[S-1, 0]]
  while len(que)>0:
    now = que.pop(0)
    for nex in edge[now[0]][now[1]]:
      if d[nex[0]][nex[1]] != -1:continue
      d[nex[0]][nex[1]] = d[now[0]][now[1]]
      if now[1] == 2:
        d[nex[0]][nex[1]] += 1
      que.append(nex)
  print(d[T-1][0])
  
if __name__ == "__main__":
  main()