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
def main():
	N, M = map(int, input().split())
	INF = 2*31-1
	to = [[] for _ in range(N)]
	for _ in range(M):
		u, v = map(int, input().split())
		to[u-1].append(v-1)
	dist = [[INF, INF, INF] for _ in range(N)]

	S, T = map(int, input().split())
	queue = []
	for pos in to[S-1]:
		queue.append([pos, 1])
		dist[pos][1] = 1
	while queue:
		pos, l = queue.pop(0)
		next_positions = to[pos]
		for next_pos in next_positions:
			nl = (l+1)%3
			if dist[next_pos][nl] != INF:
				continue
			dist[next_pos][nl] = dist[pos][l]+1
			queue.append([next_pos, nl])
	ans = dist[T-1][0]
	if ans == INF:
		print(-1)
	else:
		print(ans//3)

if __name__ == '__main__':
	main()