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
import heapq
import math

input = sys.stdin.readline
N, M = map(int, input().split())
uv = [list(map(int, input().split())) for _ in range(M)]
S, T = map(int, input().split())

g = [[] for _ in range(3*N)]
for a, b in uv:
    g[a-1].append(b-1+N)
    g[a-1+N].append(b-1+2*N)
    g[a - 1 + 2 * N].append(b - 1)

# bfsでよかった
hq = [(0, S-1)]
costs = [math.inf] * 3 * N
while len(hq) > 0:
    c, u = heapq.heappop(hq)
    if u == T - 1:
        print(c // 3)
        sys.exit()
    costs[u] = c
    for v in g[u]:
        if costs[v] > c+1:
            heapq.heappush(hq, (c + 1, v))
        costs[v] = min(costs[v], c+1)

print(-1)
