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 Python (3.4.3) (judge verdict: Runtime Error)

import numpy as np
n,m=map(int,input().split())
uv=[tuple(map(int,input().split())) for _ in range(m)]
s,t=map(int,input().split())
s,t=s-1,t-1

move=np.zeros((n,n))
for ui,vi in uv:
    ui,vi=ui-1,vi-1
    move[ui][vi]=1
# print(move)
kenkenpa=np.dot(np.dot(move,move),move)

visited=[False for _ in range(n)]
cost=[1000000007 for _ in range(n)]
prev=[None for _ in range(n)]
cost[s]=0
prev[s]=s
now=s
while True:
    min=1000000007
    next=-1
    visited[now]=True
    for i in range(n):
        if not visited[i]:
            if kenkenpa[now][i]==1:
                tmpCost=cost[now]+1
                if cost[i]>tmpCost:
                    cost[i]=tmpCost
                    prev[i]=now
            if min>cost[i]:
                min=cost[i]
                next=i
    if next==-1:
        break
    now=next
print(cost)
print(prev)
if prev[t]==None:
    print(-1)
else:
    print(cost[t])