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: Accepted)

import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;

public class Main {

    private static Scanner in = new Scanner(System.in);
    private static PrintWriter out = new PrintWriter(System.out);

    private static final BigInteger MOD = BigInteger.valueOf((long) 1e9 + 7);

    public static void main(String[] args) {
        solve();
        out.flush();
    }

    private static void solve() {
        int n = in.nextInt();
        int m = in.nextInt();
        Map<Integer, Set<Integer>> graph = new HashMap<>();

        for (int i=0; i < m; i++) {
            int u = in.nextInt();
            int v = in.nextInt();

            if (!graph.containsKey(u))
                graph.put(u, new HashSet<>());
            graph.get(u).add(v);
        }

        // out.println(graph);

        int s = in.nextInt();
        int t = in.nextInt();

        Map<Integer, Set<Integer>> graph2 = new HashMap<>();
        for (Map.Entry<Integer, Set<Integer>> e : graph.entrySet()) {
            for (int i=0; i < 3; i++) {
                if (!graph2.containsKey(e.getKey()*3 + i))
                    graph2.put(e.getKey()*3 + i, new HashSet<>());
                int finalI = i;
                e.getValue().forEach(v -> graph2.get(e.getKey()*3 + finalI).add(v*3 + (finalI +1)%3));
            }
        }

        Set<Integer> been = new HashSet<>();
        ArrayDeque<Map.Entry<Integer, Integer>> toVisit = new ArrayDeque<>();

        int ans = -1;
        if (graph2.containsKey(s*3))
            graph2.get(s*3).forEach(v -> toVisit.addLast(new AbstractMap.SimpleEntry<>(v, 1)));
        while (!toVisit.isEmpty()) {
            Map.Entry<Integer, Integer> e = toVisit.poll();
            if (been.contains(e.getKey()))
                continue;

            if (e.getKey() == t * 3) {
                ans = e.getValue() / 3;
                break;
            }
            been.add(e.getKey());
            if (graph2.containsKey(e.getKey()))
                graph2.get(e.getKey()).forEach(v -> toVisit.addLast(new AbstractMap.SimpleEntry<>(v, e.getValue() + 1)));
        }

        out.println(ans);
    }

}