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,deque
import sys,heapq,bisect,math,itertools,string,queue
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())


N,M = inpl()
lines1 = defaultdict(set)

for _ in range(M):
    u,v = inpl()
    u -= 1
    v -= 1
    lines1[u].add(v)

S,T = inpl()
S -= 1
T -= 1

#MEMO = [[set() for i in range(3)] for _ in range(N)]
#visited = [False]*N

lines2 = defaultdict(set)
def dfs(start,s,n): #s: 今いる , n: 何歩目
    if n == 3:
        lines2[start].add(s)
    else:
        for t in lines1[s]:
            dfs(start,t,n+1)

for i in range(N):
    dfs(i,i,0)

'''
print()
for s in range(N):
    for t in lines2[s]:
        print(s,t)
'''

def search (s,w_0): #s->t
    global weight
    global q

    for t in list(lines2[s]):
        w = w_0 + 1
        if weight[t] > w:
            heapq.heappush(q, [w,t])
            weight[t] = w

weight = [INF]*N
weight[S] = 0
q = [[0,S]]
heapq.heapify(q)
while q:
    w,n = heapq.heappop(q)
    search(n,w)

if weight[T] == INF:
    print(-1)
else:
    print(weight[T])