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: Wrong Answer)

import sys
import time
time0=time.time()
input = sys.stdin.readline

N,M=map(int,input().split())
E=[list(map(int,input().split())) for i in range(M)]
S,G=map(int,input().split())

ELIST=[set() for i in range(N+1)]

for x,y in E:
    ELIST[x].add(y)

check=[10**8]*(N+1)
check[S]=0

def three_step_from(S,i):

    ONE=set()

    for s in S:

        ONE=ONE|ELIST[s]

    TWO=set()
    for one in ONE:
        
        TWO=TWO|ELIST[one]

    THREE=set()
    for two in TWO:
        THREE=THREE|ELIST[two]

    RET=set()

    for three in THREE:
        if check[three]!=10**8:
            continue
        check[three]=i
        RET.add(three)

    return RET

i=1
NOW={S}
while NOW:
    NOW=three_step_from(NOW,i)
    i+=1

    if check[G]!=10**8:
        print(check[G])
        sys.exit()

    if time.time()-time0>1.8:
        print(-1)
        sys.exit()

else:
    print(-1)