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)

import sys, bisect, math, itertools, string, queue, copy
# import numpy as np
# import scipy
from collections import Counter,defaultdict,deque
from itertools import permutations, combinations
from heapq import heappop, heappush
input = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7
def inp(): return int(input())
def inpm(): return map(int,input().split())
def inpl(): return list(map(int, input().split()))
def inpls(): return list(input().split())
def inplm(n): return list(int(input()) for _ in range(n))
def inplL(n): return [list(input()) for _ in range(n)]
def inplT(n): return [tuple(input()) for _ in range(n)]
def inpll(n): return [list(map(int, input().split())) for _ in range(n)]
def inplls(n): return sorted([list(map(int, input().split())) for _ in range(n)])

def main():
    n,m = inpm()
    g = [[] for _ in range(n)]
    for _ in range(m):
        u,v = inpm()
        u -= 1
        v -= 1
        g[u].append(v)
    s,t = inpm()
    s -= 1
    t -= 1
    que = deque()
    que.append(s)
    visited1 = [0 for _ in range(n)]
    visited2 = [0 for _ in range(n)]
    visited3 = [0 for _ in range(n)]
    visited3[s] = 1
    cnt = 1
    while True:
        tank1 = deque()
        tank2 = deque()
        tank3 = deque()
        while que:
            start = que.pop()
            for go in g[start]:
                if not visited1[go]:
                    visited1[go] = 1
                    tank1.append(go)
        while tank1:
            start = tank1.pop()
            for go in g[start]:
                if not visited2[go]:
                    visited2[go] = 1
                    tank2.append(go)
        while tank2:
            start = tank2.pop()
            for go in g[start]:
                if not visited3[go]:
                    visited3[go] = 1
                    tank3.append(go)
                if go == t:
                    print(cnt)
                    return
        que = tank3
        cnt += 1
        if que == deque():
            break
    print(-1)
    
if __name__ == "__main__":
    main()