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

sys.setrecursionlimit(10000000)

n,m = map(int,input().split())
to=[[] for _ in range(n)]
dist=[[0] *3 for _ in range(100005)]
INF=10**15

for i in range(m):
    a,b=map(int,input().split())
    a-=1;b-=1
    to[a].append(b)

sv,tv=map(int,input().split())
sv-=1;tv-=1

for i in range(n):
    for j in range(3):
        dist[i][j] =INF

q=deque()
q.append((sv,0))
dist[sv][0]=0

while q:
    v,l=q.popleft()
    for u in to[v]:
        nl =(l+1)%3
        if dist[u][nl] != INF: continue
        dist[u][nl] = dist[v][l]+1
        q.append((u,nl))

ans=dist[tv][0]
if ans == INF:ans=-1
else: ans //=3
print(ans)