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.Text;
using System.Linq;
using static MyIO;


public class E
{
	public static void Main()
	{
		int N = GetInt();
		int M = GetInt();
		var adj = new List<int>[N];
		for(int i = 0; i < N; i++)
			adj[i] = new List<int>();
		for(int i = 0; i < M; i++)
			adj[GetInt() - 1].Add(GetInt() - 1);
		int S = GetInt() - 1;
		int T = GetInt() - 1;

		long[,] d = new long[3, N];
		for(int i = 0; i < 3; i++)
			for(int j = 0; j < N; j++)
				d[i, j] = -1;

		d[0, S] = 0;
		var queue = new Queue<Tuple<int, int>>();
		queue.Enqueue(Tuple.Create(0, S));
		while(queue.Count > 0)
		{
			var e = queue.Dequeue();
			int state = e.Item1;
			int v = e.Item2;
			foreach(int u in adj[v])
			{
				int nextState = (state + 1) % 3;
				if(d[nextState, u] != -1)
					continue;

				d[nextState, u] = d[state, v] + 1;
				queue.Enqueue(Tuple.Create(nextState, u));
			}
		}

		Console.WriteLine((d[0, T] == -1) ? -1: d[0, T] / 3);
	}
}


public static class MyIO
{
	private static string[] args = null;
	private static int num = -1;
	private static int used = -1;

	private static string getArg()
	{
		if(used == num)
		{
			args = Console.ReadLine().Split(' ');
			num = args.Length;
			used = 0;
		}
		return args[used++];
	}

	public static int GetInt(){ return int.Parse(getArg()); }
	public static long GetLong(){ return long.Parse(getArg()); }
	public static double GetDouble(){ return double.Parse(getArg()); }
	public static string GetString(){ return getArg(); }
}