Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from collections import defaultdict
n, m = list(map(int, input().split()))
d1 = defaultdict(list)
for i in range(m):
  u, v = list(map(int, input().split()))
  d1[u].append(v)
s, t = list(map(int, input().split()))
d2 = defaultdict(set)

def dfs (x, nx, cnt):
  if cnt==3:
    d2[x].add(nx)
  elif cnt>3:
    return

  for nnx in d1[nx]:
    dfs(x, nnx, cnt+1)

for i in range(1, n+1):
  dfs(i, i, 0)

from heapq import heappop, heappush
prev = [-1 for _ in range(n+1)]
def dijkstra(s, graph):
  d = [float("INF") for _ in range(n+1)]
  d[s] = 0
  q = []
  heappush(q, [0, s])

  while q:
    dt, v = heappop(q)
    if d[v] < dt:
      continue
    for to in graph[v]:
      if d[to] > dt+1:
        d[to] = dt+1
        prev[to] = v
        heappush(q, [d[to], to])
  return d
dd = dijkstra(s, d2)
if dd[t]==float("INF"):
  print(-1)
else:
  print(dd[t])