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
# 木構造を作る
N,M = map(int, input().split())

G = [[] for i in range(3*N)]
for i in range(M):
    # グラフに頂点を追加(距離があるときは,u,vの後に入れる)
    u,v = map(int,input().split())
    G[3*(u-1)].append((3*(v-1)+1))
    G[3*(u-1)+1].append((3*(v-1)+2))
    G[3*(u-1)+2].append((3*(v-1)))

S,T = map(int, input().split())

# 木をBFSをする
dist = [-1] * (3*N)
dist[3*(S-1)] = 0
q = deque([3*(S-1)])
while q:
    v = q.popleft()
    d = dist[v]
    Vs = G[v]
    for u in Vs:
        if dist[u] == -1:
            q.append(u)
            dist[u] = d + 1

ans = dist[3*(T-1)]
print(-1 if ans == -1 else ans//3)
