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())
w = [set() for _ in range(n + 1)]
for _ in range(m):
  u, v = map(int, input().split())
  w[u].add(v)
s, t = map(int, input().split())
l = [[-1, -1, -1] for _ in range(n + 1)] # 頂点までの最短移動回数（余りごとに３つ）

d = deque()
d.append((s, 0, 0))
while d: # qが小さい順に制御されている
  i, q, r = d.popleft() # 頂点番号、商、余り
  for _w in w[i]:
    if _w == t and r == 2: # 終了条件
      print((q + 1) // 3)
      exit()
    if l[_w][(r + 1) % 3] == -1: # 余り考慮での初到達時に、メモとTODO追加
      l[_w][(r + 1) % 3] = q + 1
      d.append((_w, q + 1, (r + 1) % 3))
print(-1)