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)

import sys
n,m = map(int,sys.stdin.readline().split())
# sys.stdout.write('{} {}'.format(n,m))
vertices = [[] for _ in range(n+1)]
jumpValue = [[-1,-1,-1] for _ in range(n+1)]
visited = set()
queue = []
def bfs(arr,jump):
    temp = []
    for node in arr:
        # print('NODE:',node,jumpValue[node][(jump+1)%3])
        if jumpValue[node][(jump+1)%3]==-1:
            queue.append([node,jump+1])
            # queue.append(node)
        temp = temp+vertices[node]
        # if node==t:
        #     print('YES',jump+1)
        #     jumpValue[node].append(jump+1)
    return temp
for i in range(m):
    a,b = map(int,sys.stdin.readline().split())
    vertices[a].append(b)
s,t = map(int,sys.stdin.readline().split())
visited.add(s)
queue.append([s,0])
# print(vertices)
done =set()
while(len(queue)>0 ):
    next=queue.pop(0)
    if jumpValue[next[0]][next[1]%3] == -1:
        jumpValue[next[0]][next[1]%3] = next[1]
        # print('Next:', next[0])
        # print(jumpValue)
        # print('Queue:', queue)
        # print('NEXT', next, vertices[next[0]])
        run = bfs(vertices[next[0]], next[1])
        run = bfs(run, next[1] + 1)
        bfs(run, next[1] + 2)
        # print('----------------------------------------------')
print(max(jumpValue[t][0]//3,-1))