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: Time Limit Exceeded)

import java.util.*;

public class Main {

	private static List<Integer>[] g;
	private static boolean[] visited;
	private static Set<Integer>[] ng;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();

		g = new ArrayList[n];
		visited = new boolean[n];
		ng = new HashSet[n];
		for (int i = 0; i < n; i++) {
			g[i] = new ArrayList<>();
			ng[i] = new HashSet<>();
		}

		for (int i = 0; i < m; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			g[u].add(v);
			//g[v].add(u);
		}

		int s = sc.nextInt() - 1;
		int t = sc.nextInt() - 1;

		dfs(s, s, 0);

		if (!visited[t]) {
			System.out.println(-1);
			return;
		}

		//for (int i = 0; i < ng.length; i++) {
		//	System.out.print(i);
		//	System.out.print(": ");
		//	for (Integer v : ng[i]) {
		//		System.out.print(v);
		//		System.out.print(" ");
		//	}
		//	System.out.println();
		//}

		Queue<int[]> q = new LinkedList<>();
		q.offer(new int[]{s, 0});
		while (!q.isEmpty()) {
			int[] v = q.poll();
			if (v[0] == t) {
				System.out.println(v[1]);
				return;
			}
			for (Integer i : ng[v[0]]) {
				q.offer(new int[]{i, v[1] + 1});
			}
		}
	}

	private static void dfs(int v, int p, int c) {
		if (c == 3) {
			ng[p].add(v);
			//ng[v].add(p);
			c = 0;
		}
		if (c == 0) {
			if (visited[v]) return;
			visited[v] = true;
		}

		for (int i = 0; i < g[v].size(); i++) {
			dfs(g[v].get(i), (c == 0) ? v : p, c + 1);
		}
	}
}