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)

def main():
    import sys
    input = sys.stdin.readline
    sys.setrecursionlimit(10**7)
    from collections import Counter, deque
    #from collections import defaultdict
    from itertools import combinations, permutations, accumulate, groupby, product
    from bisect import bisect_left,bisect_right
    from heapq import heapify, heappop, heappush
    import math

    inf = 10**17
    #mod = 10**9 + 7

    n,m = map(int, input().split())
    adj = [[] for _ in range(n)] #頂点数, 場合によって変える
    for _ in range(m):
        a,b = map(int, input().split())
        adj[a-1].append(b-1)
    s,t = map(int, input().split())
    s -= 1
    t -= 1

    d1 = [inf]*n
    d2 = [inf]*n
    d0 = [inf]*n
    d0[s] = 0
    d = deque([[s, 0]])
    while d:
        v, hosuu = d.popleft()
        hosuu += 1
        for nv in adj[v]:
            if hosuu%3==1 and d1[nv]>hosuu:
                d1[nv] = hosuu 
                d.append([nv, hosuu])
            if hosuu%3==2 and d2[nv]>hosuu:
                d2[nv] = hosuu
                d.append([nv, hosuu])
            if hosuu%3==0 and d0[nv]>hosuu:
                d0[nv] = hosuu
                d.append([nv, hosuu])
    if d0[t]==inf:
        print(-1)
    else:
        print(d0[t]//3)

if __name__ == '__main__':
    main()