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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;
public class Main {
  
  private static long X = 1000000007;
  
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    int[][] graph = new int[N+1][N+1];
    for (int i = 0; i < M; i++) {
      int u = sc.nextInt();
      int v = sc.nextInt();
      graph[u][v] = 1;
    }
    int S = sc.nextInt();
    int T = sc.nextInt();
    
    Set<Integer> reachable = new HashSet<>();
    reachable.add(T);
    boolean hasNewNode = true;
    
    boolean reachStart = false;
    int count = 0;
    
    LinkedList<Integer> queue = new LinkedList<>();
    queue.add(T);
    while (hasNewNode && !reachStart) {
      
      check(graph, queue);
      count++;
      
      hasNewNode = false;
      Iterator<Integer> itr = queue.iterator();
      while(itr.hasNext()){
        int node = itr.next();
        if (node == S) {
          reachStart = true;
        }
        hasNewNode |= reachable.add(node);
      }
    }
    
    if (reachStart) {
      System.out.println(count);
    } else {
      System.out.println(-1);
    }
  }
  
  private static void check(int[][] graph, LinkedList<Integer> queue) {

    for (int c = 0; c < 3; c++) {
      int nodeCount = queue.size();
      while (nodeCount-- > 0) {
        Integer node = queue.remove();
        for (int i = 1; i < graph.length; i++) {
          int v = graph[i][node];
          if (v == 1) {
            queue.add(i);
          }
        }
      }
    }
  }
}