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
from collections import deque
mod = 10**9 + 7

n, m = [int(x) for x in input().split()]
g = [[] for _ in range(n)]

for _ in range(m):
    u, v = [int(x) for x in input().split()]
    g[u - 1].append(v - 1)

s, t = [int(x) for x in input().split()]

d = [[-1, -1, -1] for _ in range(n)]

def bfs(u):
    queue = deque([[u, 0]])
    d[u][0] = 0
    while queue:
        v, d_ = queue.popleft()
        for i in g[v]:
            if d[i][(d_ + 1) % 3] < 0:
                d[i][(d_ + 1) % 3] = d_ + 1
                queue.append([i, d_ + 1])
    return d

d = bfs(s - 1)

print(d[t - 1][0] // 3)
