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: Wrong Answer)

import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		ArrayList<Integer>[] list = new ArrayList[n+1];
		for(int i = 1; i <= n; i++){
		    list[i] = new ArrayList<Integer>();
		}
		for(int i = 0; i < m; i++){
		    int from = sc.nextInt();
		    int to = sc.nextInt();
		    list[from].add(to);
		}
		int s = sc.nextInt();
		int t = sc.nextInt();

		// 計算
		int result = -1;
		PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
		int[] firsts = new int[n+1];
		Arrays.fill(firsts, Integer.MAX_VALUE);
		int[] seconds = new int[n+1];
		Arrays.fill(seconds, Integer.MAX_VALUE);
		int[] thirds = new int[n+1];
		Arrays.fill(thirds, Integer.MAX_VALUE);
		pq.add(s);
		thirds[s] = 0;
		while(!pq.isEmpty()){
		    Integer now = pq.poll();
		    ArrayList<Integer> first = list[now];
		    for(Integer i : first){
		        if(firsts[i] <= thirds[now]) continue;
		        firsts[i] = thirds[now];
		        ArrayList<Integer> second = list[i];
		        for(Integer j : second){
		            if(seconds[i] <= thirds[now]) continue;
		            seconds[i] = thirds[now];
		            ArrayList<Integer> third = list[j];
		            for(Integer k : third){
		                int before = thirds[k];
		                thirds[k] = min(thirds[k], thirds[now] + 1);
		                if(thirds[k] != before) pq.add(k);
		            }
		        }
		    }
		}
		if(thirds[t] != Integer.MAX_VALUE) result = thirds[t];

		// 出力
		System.out.println(result);

	}
}