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)

#!/usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

#A
def A():
    s = S()
    b = list(set(s))
    a = len(list(set(s)))
    if a == 2 and s.count(b[0]) == s.count(b[1]) == 2:
        print("Yes")
    else:
        print("No")
    return

#B
def B():
    n = II()
    p = LI()
    ans = 0
    for i in range(1,n-1):
        if (p[i - 1] < p[i] and p[i] < p[i + 1]) or (p[i + 1] < p[i] and p[i] < p[i - 1]):
            ans += 1
    print(ans)
    return

#C
def C():
    n = II()
    d = LI()
    d.sort()
    ans = 0
    for i in range(10**5+1):
        if n // 2 == bisect_left(d, i):
            ans += 1
    print(ans)
    return


def combination_mod(n, k, mod):
    # power_funcを用いて(nCk) mod p を求める 
    # nCk = n!/((n-k)!k!)を使用

    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    a = factorial(n) % mod
    b = factorial(k) % mod
    c = factorial(n - k) % mod
    return (a * power_func(b, mod - 2, mod) * power_func(c, mod - 2, mod)) % mod
    
def power_func(a, b, mod):
    # a^b mod p を求める
    # bを2進数分解して高速累乗

    if b == 0: return 1
    if b % 2 == 0:
        d = power_func(a, b // 2, mod)
        return d * d % mod
    if b % 2 == 1:
        return (a * power_func(a, b - 1, mod)) % mod

#D
def D():
    n, k = LI()
    ans = 0
    for i in range(1,k+1):
        a = combination_mod(n - k + 1, i, mod)
        b = combination_mod(k - 1, i - 1, mod)
        print((a*b)%mod)
    return

def dijkstra(num, start, edge, g):

    dist = defaultdict(lambda: inf)
    dist[(start,0)] = 0
    q = [(dist[(start,0)], start, 0)]
    while q:
        du, u, t = heappop(q)
        for j in edge[u]:
            if t > 3:
                t = 0
            if dist[(j, t + 1)] > du + 1:
                dist[(j, t + 1)] = du + 1
                heappush(q, [dist[(j, t + 1)], j, t + 1])
    return dist
#E
def E():
    n, m = LI()
    v = LIR_(m)
    s, t = LI_()
    edge = defaultdict(list)
    for x,y in v:
        edge[x].append(y)
    d = dijkstra(n, s, edge, t)
    if d[(t,2)] == inf:
        print(-1)
    else:
        print(d[(t,2)])
    return

#F
def F():
    return

#G
def G():
    return

#H
def H():
    return

#Solve
if __name__ == '__main__':
    E()