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 {
    static ArrayList<Integer>[] graph;
    static int n;
    static int[] visited;
	public static void main (String[] args)  {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		graph = new ArrayList[3 * n];
		for (int i = 0; i < 3 * n; i++) {
		    graph[i] = new ArrayList<Integer>();
		}
		int m = sc.nextInt();
		for (int i = 0; i < m; i++) {
		    int a = sc.nextInt() - 1;
		    int b = sc.nextInt() - 1;
		    graph[a].add(b + n);
		    graph[a + n].add(b + 2 * n);
		    graph[a + 2 * n].add(b);
		    
		}
		int s = sc.nextInt() - 1;
		int t = sc.nextInt() - 1;
		visited = new int[n * 3];
		search(s, 1);
		if (visited[t] == 0) {
		    System.out.println(-1);
		} else {
	    	System.out.println(visited[t] / 3);
		}
	}
	
	static void search(int idx, int value) {
	    if (visited[idx] != 0 && visited[idx] <= value) {
	        return;
	    }
	    visited[idx] = value;
	    for (int x : graph[idx]) {
	        search(x, value + 1);
	    }
	}
}