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

N,M=map(int,input().split())
E=[list(map(int,input().split())) for i in range(M)]
S,G=map(int,input().split())

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

for x,y in E:
    ELIST[x].append(y)

DIS=[[-1,-1,-1] for i in range(N+1)]
DIS[S][0]=0

from collections import deque

S=deque([S])
i=1

while S:
    #print(S)
    NEWS=deque()
    while S:     
        x=S.pop()

        for to in ELIST[x]:
            if DIS[to][i%3]==-1:
                DIS[to][i%3]=i
                NEWS.append(to)

    i+=1
    S=NEWS

print(DIS[G][0]//3)
            
        
    
