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
from collections import deque

N,M=map(int, sys.stdin.readline().split())

nl=[ [] for i in range(N+1) ]

for i in range(M):
	u,v=map(int, sys.stdin.readline().split())
	nl[u].append(v)

S,T=map(int, sys.stdin.readline().split())

V=[ [ None for j in range(3) ] for i in range(N+1) ]
V[S][0]=0


q=deque()
q.append( (S,0) )

while q:
	x=q.popleft()
	fro=x[0]
	cur_mod=x[1]

	for to in nl[fro]:
		next_mod=(cur_mod+1)%3
		if V[to][next_mod] is None:
			V[to][next_mod]=V[fro][cur_mod]+1
			if V[T][0] is not None:
				break
			q.append( ( to, next_mod ) )


if V[T][0] is None:
	print -1
else:
	print V[T][0]/3