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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader r = new BufferedReader(new InputStreamReader(System.in), 1);
		String s;
		String sl[];

		s = r.readLine();
		sl = s.split(" ");

		int n = Integer.parseInt(sl[0]);
		int m = Integer.parseInt(sl[1]);

		ArrayList<ArrayList<Integer>> g = new ArrayList<>(300000);
		boolean b[] = new boolean[300000];

		for (int i = 0; i < 3 * n; i++) {
			g.add(new ArrayList<>());
		}

		for (int i = 0; i < m; i++) {
			s = r.readLine();
			sl = s.split(" ");
			int u = Integer.parseInt(sl[0]) - 1;
			int v = Integer.parseInt(sl[1]) - 1;
			g.get(u).add(v + n);
			g.get(u + n).add(v + 2 * n);
			g.get(u + 2 * n).add(v);
		}

		// System.out.println(g);
		s = r.readLine();
		sl = s.split(" ");

		int start = Integer.parseInt(sl[0]) - 1;
		int term = Integer.parseInt(sl[1]) - 1;

		ArrayDeque<Integer> qv = new ArrayDeque<>();
		ArrayDeque<Integer> qd = new ArrayDeque<>();

		qv.offer(start);
		qd.offer(0);

		while (!qv.isEmpty()) {
			int v = qv.poll();
			int d = qd.poll();
			// System.out.println(v + ", " + d);
			if (v == term) {
				System.out.println(d / 3);
				System.exit(0);
			}
			for (Integer e : g.get(v)) {
				if (!b[e]) {
					qv.offer(e);
					qd.offer(d + 1);
					b[e] = true;
				}
			}
		}
		System.out.println(-1);
	}

}