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 = map(int, input().split())
link = [[] for _ in range(n)]
for i in range(m):
    tmp = list(map(int,input().split()))
    link[tmp[0]-1].append(tmp[1]-1)
s,t = map(int, input().split())

from collections import deque
Q = deque()
Q.append([s-1,0,0])
visited = [[] for _ in range(n)]
visited[s-1]=[0]

while Q:
    now,kkp,cnt = Q.popleft()
    nxt_kkp = (kkp+1)%3
    for nxt in link[now]:
        if nxt == (t-1) and nxt_kkp == 0:
            print((cnt+1)//3)
            exit()
        if nxt_kkp in visited[nxt]:
            continue
        visited[nxt].append(nxt_kkp)
        Q.append([nxt,nxt_kkp,cnt+1])

print(-1)