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 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int numNode = sc.nextInt();
		int numEdge = sc.nextInt();
		ArrayList<ArrayList<Integer>> go = new ArrayList<>();
		for(int i = 0; i < numNode; i ++) {
			go.add(new ArrayList<Integer>());
		}
		for(int i = 0; i < numEdge; i ++) {
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			go.get(a).add(b);
		}
		int start = sc.nextInt() - 1;
		int goal = sc.nextInt() - 1;

		long need[] = new long[numNode];
		for(int i = 0; i < numNode; i ++) {
			need[i] = -100;
		}

		Deque<Integer> stack = new ArrayDeque<Integer>();
		stack.addFirst(start);

		long count = 0;
		while (!stack.isEmpty()) {
			int size = stack.size();
			for(int a = 0; a < size; a ++) {
				int crt = stack.removeFirst();
				if(need[crt] >= 0 && count % 3 == 0) { continue; }
				if(need[crt] == - count % 3) { continue; }
				if(count % 3 == 0) {
					need[crt] = count / 3;
				}else {
					need[crt] = - count % 3;
				}
				for(int i = 0; i < go.get(crt).size(); i ++) {
					stack.addLast(go.get(crt).get(i));
				}
			}
			if(need[goal] >= 0) { break; }
			count ++;
		}

		System.out.println(need[goal] == -100 ? -1 : need[goal]);
	}
}