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 sys
from collections import deque

def main():
  input = sys.stdin.readline
  n, m = map(int, input().split())
  roads0 = [[] for _ in range(n)]
  roads1 = [[] for _ in range(n)]
  roads2 = [[] for _ in range(n)]
  
  for i in range(m):
    u, v = map(int, input().split())
    roads0[u-1].append(v-1)
    roads1[u-1].append(v-1)
    roads2[u-1].append(v-1)
    
  ans = 100000000
  
  s, t = map(int, input().split())
  not_yet = deque([])
  already = [[False]*3 for _ in range(n)]
  dist = [[0]*3 for _ in range(n)]
  for v in roads0[s-1]:
    not_yet.append((v, 1))
    dist[v][1] = 1
    already[v][1] = True
  already[s-1][0] = True
  while not_yet:
    key = not_yet.pop()
    k1, k2 = key[0], key[1]
    if k2 == 0:
      for v in roads1[k2]:
        if already[v][1]:
          continue
        dist[v][1] = dist[k1][0]+1
        already[v][1] = True
        not_yet.append((v, 1))
    elif k2 == 1:
      for v in roads2[k2]:
        if already[v][2]:
          continue
        dist[v][2] = dist[k1][1]+1
        already[v][2] = True
        not_yet.append((v, 2))
    else:
      for v in roads3[k2]:
        if already[v][0]:
          continue
        dist[v][0] = dist[k1][2]+1
        already[v][0] = True
        not_yet.append((v, 0))
  if already[t-1][0]:
    print(dist[t-1][0])
  else:
    print(-1)

    
if __name__ == "__main__":
  main()