Solve in Python:
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-----
4 4
1 2
2 3
3 4
4 1
1 3

-----Sample Output-----
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.
from collections import deque 
MOD = 10**9 +7
INF = 10**9
def main():
    n,m = list(map(int,input().split()))
    V0 = [[] for _ in range(n)]
    V1 = [[] for _ in range(n)]
    V2 = [[] for _ in range(n)]
    for _ in range(m):
        u,v = list(map(int,input().split()))
        u -= 1
        v -= 1
        V0[u].append(v)
        V1[u].append(v)
        V2[u].append(v)
    
    s,t = list(map(int,input().split()))
    s -= 1
    t -= 1
    
    d0 = [-1]*n
    d1 = [-1]*n
    d2 = [-1]*n
    d0[s] = 0

    q = deque()
    q.append((s,0))
    while len(q):
        v,mod = q.popleft()
        if mod == 0:
            for e in V0[v]:
                if d1[e] == -1:
                    d1[e] = d0[v] + 1
                    q.append([e,1])

        elif mod == 1:
            for e in V1[v]:
                if d2[e] == -1:
                    d2[e] = d1[v] + 1
                    q.append([e,2])
        
        else:
            for e in V2[v]:
                if e == t:
                    print(((d2[v] + 1)//3))
                    return           
                if d0[e] == -1:
                    d0[e] = d2[v] + 1
                    q.append([e,0])
    
    if d0[t] == -1:
        print((-1))
   
def __starting_point():
    main()

__starting_point()