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
import sys
input = sys.stdin.readline
inpl = lambda: list(map(int,input().split()))

N, M = inpl()
uv = [ set() for _ in range(3*N) ]
vu = [ set() for _ in range(3*N) ]
for _ in range(M):
    u, v = inpl()
    for k in range(3):
        uv[u-1 + k*N].add(v-1 + ((k+1)%3)*N)
        vu[v-1 + ((k+1)%3)*N].add(u-1 + k*N)
S, T = inpl()
S -= 1
T -= 1

cur = set({S})
for i in range(3*N):
    if i % 3 == 0 and T in cur:
        print(i//3)
        break
    elif len(cur) == 0:
        print(-1)
        break
    prev = cur
    cur = set()
    for u in prev:
        for fr in vu[u]:
            uv[fr].remove(u)
        cur |= uv[u]
else:
    print(-1)