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: Accepted)

import sys
from fractions import gcd
from collections import defaultdict
from collections import deque
sys.setrecursionlimit(10**7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def LI(): return list(map(int, sys.stdin.readline().split()))
def II(): return int(sys.stdin.readline())
def LS(): return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]

def main():
    n, m = LI()
    g = [[] for i in range(3*(n+1))]
    for i in range(m):
        u, v = LI()
        g[u].append(v + n + 1)
        g[u + n + 1].append(v + 2*(n + 1))
        g[u + 2*(n + 1)].append(v)

    s, t = LI()

    Q = deque([s])
    searched = [0 for i in range(3*(n+1))]
    searched[s] = 1
    depth = [0 for i in range(3*(n+1))]
    while len(Q) != 0:
        v = Q.popleft()
        for i in g[v]:
            if i == t:
                depth[i] = depth[v] + 1
                print(depth[i]//3)
                return
            if searched[i] == 0:
                depth[i] = depth[v] + 1
                Q.append(i)
                searched[i] = 1
    print(-1)


if __name__ == "__main__":
    main()