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: Wrong Answer)

/**
 * Created at 12:39 on 2019-06-30
 */

import java.io.*;
import java.util.*;

public class Main {

  static FastScanner sc = new FastScanner();
  static PrintStream out = System.out;
  static PrintWriter pw = new PrintWriter(out);

  static final int[] dx = {0, 1, 0, -1};
  static final int[] dy = {-1, 0, 1, 0};

  static final long MOD = (long) (1e9 + 7);
  static final long INF = Long.MAX_VALUE / 2;

  public static class Solver extends Graph {
    public Solver() {

      int N = sc.nextInt();
      int M = sc.nextInt();

      read(N, M, false, -1);

      int S = sc.nextInt()-1;
      int T = sc.nextInt()-1;

      ArrayDeque<Node> queue = new ArrayDeque();

      queue.add(nodes[S*3]);
      nodes[S*3].cost = 0;

      while (!queue.isEmpty()) {
        Node from = queue.poll();
        for(Edge e: from.edgesOut) {
          Node to = e.getTo();
          if (to.cost == -1) {
            to.cost = from.cost + 1;
            queue.add(to);
            if (to.id == T*3) {
              queue.clear();
              break;
            }
          }
        }
      }

      out.println(nodes[T*3].cost == -1 ? -1 : nodes[T*3].cost / 3);

    }
  }

  public static class Graph {

    Node[] nodes;

    /*
     * N: 頂点の数
     * M: 辺の数
     * hasCost: 辺にコストがあるか
     */
    public void read(int N, int M, boolean hasCost, long defaultCost) {
      nodes = new Node[N*3];
      for (int i=0; i<N; i++) {
        for (int j=0; j<3; j++) {
          nodes[i * 3 + j] = new Node(i, defaultCost, j);
        }
      }

      for (int i=0; i<M; i++) {
        int from = sc.nextInt()-1;
        int to = sc.nextInt()-1;
        long cost = (hasCost ? sc.nextLong() : 1);
        for (int j=0; j<3; j++) {
          Edge e = new Edge(from*3 + j, to*3 + ((j+1)%3), cost);
          nodes[from*3+j].edgesOut.add(e);
          nodes[to*3+((j+1)%3)].edgesIn.add(e);
        }
      }
    }

    public void init() {
      for (int i=0; i<nodes.length; i++) {
        nodes[i].init();
      }
    }

    public void reverse() {
      for (Node n: nodes) {
        n.reverse();
      }
    }

    public class Node {
      int id;
      long cost, defaultCost;
      ArrayList<Edge> edgesIn = new ArrayList(); //このノードを終点とする辺
      ArrayList<Edge> edgesOut = new ArrayList(); //このノードを始点とする辺
      boolean done = false;
      int state;
      public Node(int id, long cost, int state) {
        this.id = id;
        this.cost = defaultCost = cost;
        this.state = state;
      }
      public void init() {
        cost = defaultCost;
        done = false;
      }
      public long cost() {
        return cost;
      }
      public void reverse() {
        ArrayList<Edge> temp = edgesIn;
        edgesIn = edgesOut;
        edgesOut = temp;
        for (Edge e: edgesIn) {
          if (e.to != this.id) {
            e.from = e.to;
            e.to = this.id;
          }
        }
        for (Edge e: edgesOut) {
          if (e.from != this.id) {
            e.to = e.from;
            e.from = this.id;
          }
        }
      }
    }

    public class Edge {
      int from, to;
      long cost;
      public Edge(int from, int to, long cost) {
        this.from = from;
        this.to = to;
        this.cost = cost;
      }
      public Node getFrom() {
        return nodes[from];
      }
      public Node getTo() {
        return nodes[to];
      }
    }

  }


  public static void main(String[] args) {
    new Solver();
  }

  static class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = in.read(buffer);
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte()) return buffer[ptr++];
      else return -1;
    }

    private static boolean isPrintableChar(int c) {
      return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
      while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
      skipUnprintable();
      return hasNextByte();
    }

    public String next() {
      if (!hasNext()) throw new NoSuchElementException();
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while (isPrintableChar(b)) {
        sb.appendCodePoint(b);
        b = readByte();
      }
      return sb.toString();
    }

    public long nextLong() {
      if (!hasNext()) throw new NoSuchElementException();
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
        minus = true;
        b = readByte();
      }
      if (b < '0' || '9' < b) {
        throw new NumberFormatException();
      }
      while (true) {
        if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
        } else if (b == -1 || !isPrintableChar(b)) {
          return minus ? -n : n;
        } else {
          throw new NumberFormatException();
        }
        b = readByte();
      }
    }

    public int nextInt() {
      return (int) nextLong();
    }

    public int[] nextIntArray(int N, boolean oneBased) {
      if (oneBased) {
        int[] array = new int[N + 1];
        for (int i = 1; i <= N; i++) {
          array[i] = sc.nextInt();
        }
        return array;
      } else {
        int[] array = new int[N];
        for (int i = 0; i < N; i++) {
          array[i] = sc.nextInt();
        }
        return array;
      }
    }

    public long[] nextLongArray(int N, boolean oneBased) {
      if (oneBased) {
        long[] array = new long[N + 1];
        for (int i = 1; i <= N; i++) {
          array[i] = sc.nextLong();
        }
        return array;
      } else {
        long[] array = new long[N];
        for (int i = 0; i < N; i++) {
          array[i] = sc.nextLong();
        }
        return array;
      }
    }
  }

}