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.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		List<Integer>[] g = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			g[i] = new ArrayList<>();
		}
		for (int i = 0; i < m; i++) {
			int v = sc.nextInt() - 1;
			int u = sc.nextInt() - 1;
			g[v].add(u);
		}
		int s = sc.nextInt() - 1;
		int t = sc.nextInt() - 1;
		int[][] d = new int[n][3];
		for (int i = 0; i < d.length; i++) {
			Arrays.fill(d[i], Integer.MAX_VALUE);
		}

		Queue<int[]> q = new LinkedList<>();
		q.offer(new int[]{s, 0});
		d[s][0] = 0;

		while (!q.isEmpty()) {
			int[] vx = q.poll();
			int v = vx[0];
			int state = vx[1];
			int next = (vx[1] + 1) % 3;
			for (int i = 0; i < g[v].size(); i++) {
				if (d[g[v].get(i)][next] != Integer.MAX_VALUE) continue;
				d[g[v].get(i)][next] = d[v][state] + 1;
				q.offer(new int[]{g[v].get(i), next});
			}
		}

		if (d[t][0] == Integer.MAX_VALUE) {
			System.out.println(-1);
			return;
		}

		System.out.println(d[t][0] / 3);
	}
}