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
N, M = [int(i) for i in input().split()]
E = [[] for _ in range(3 * N)]
for _ in range(M):
    u, v = [int(i) - 1 for i in input().split()]
    E[3 * u].append(3 * v + 1)
    E[3 * u + 1].append(3 * v + 2)
    E[3 * u + 2].append(3 * v)
S, T = [3 * (int(i) - 1) for i in input().split()]

from collections import deque
Q = deque()
Q.append((S, 0))

visited = [False] * (3 * N)
visited[S] = True

ans = -1
while Q:
    a, d = Q.popleft()
    if a == T:
        ans = d
        break
    for b in E[a]:
        if not visited[b]:
            Q.append((b, d + 1))
            visited[b] = True

if ans % 3 == 0:
    print(ans // 3)
else:
    print(-1)
