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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class TEST{
	static void Main(){
		Sol mySol =new Sol();
		mySol.Solve();
	}
}

class Sol{
	public void Solve(){
		
		int[] bw = new int[N];
		List<int>[] E = new List<int>[N];
		for(int i=0;i<N;i++){
			E[i] = new List<int>();
		}
		for(int i=0;i<M;i++){
			E[U[i]].Add(V[i]);
		}
		
		int[][] min = new int[3][];
		for(int i=0;i<3;i++){
			min[i] = new int[N];
			for(int j=0;j<N;j++){
				min[i][j] = -3;
			}
		}
		
		var q = new Queue<int>();
		min[0][S] = 0;
		q.Enqueue(0 * N + S);
		while(q.Count > 0){
			var np = q.Dequeue();
			int t = np / N;
			int now = np % N;
			int nt = (t + 1) % 3;
			foreach(var nxt in E[now]){
				if(min[nt][nxt] == -3){
					min[nt][nxt] = min[t][now] + 1;
					q.Enqueue(nt * N + nxt);
				}
			}
		}
		
		Console.WriteLine(min[0][T] / 3);
		
		
	}
	int N,M;
	int[] U,V;
	int S,T;
	public Sol(){
		var d = ria();
		N = d[0]; M = d[1];
		U = new int[M];
		V = new int[M];
		for(int i=0;i<M;i++){
			d = ria();
			U[i] = d[0] - 1;
			V[i] = d[1] - 1;
		}
		d = ria();
		S = d[0] - 1; T = d[1] - 1;
	}

	static String rs(){return Console.ReadLine();}
	static int ri(){return int.Parse(Console.ReadLine());}
	static long rl(){return long.Parse(Console.ReadLine());}
	static double rd(){return double.Parse(Console.ReadLine());}
	static String[] rsa(char sep=' '){return Console.ReadLine().Split(sep);}
	static int[] ria(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>int.Parse(e));}
	static long[] rla(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>long.Parse(e));}
	static double[] rda(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>double.Parse(e));}
}