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: Wrong Answer)

using System;
using System.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;

namespace AtCorder
{
    class Program
    {
        static void Main(string[] args)
        {
            new Program().Solve(new ConsoleInput(Console.In, ' '));
        }
        public void Solve(ConsoleInput cin)
        {
            var n = cin.ReadInt;
            var m = cin.ReadInt;
            int[] u = new int[m];
            int[] v = new int[m];
            for (int i = 0; i < m; i++)
            {
                u[i] = cin.ReadInt;
                v[i] = cin.ReadInt;
            }
            var s = cin.ReadInt;
            var t = cin.ReadInt;
            var Q = new Queue<int>();
            Q.Enqueue(s);
            Q.Enqueue(-1);
            int now= 0;
            int step = 0;
            while (Q.Count != 0)
            {
                now = Q.Dequeue();
                if(now == -1)
                {
                    step++;
                    Q.Enqueue(-1);
                    continue;
                }
                if(step != 0 && step % 3 == 0)
                {
                    if(now == t)
                    {
                        WriteLine(step / 3);
                        return;
                    }
                }
                for(int i = 0; i < m; i++)
                {
                    if(now == u[i])
                    {
                        Q.Enqueue(v[i]);
                    }
                }
                if(step > 10000)
                {
                    WriteLine(-1);
                    return;
                }
            }
            WriteLine(-1);
        }
    }

    public class ConsoleInput
    {
        private readonly System.IO.TextReader _stream;
        private char _separator = ' ';
        private Queue<string> inputStream;
        public ConsoleInput(System.IO.TextReader stream, char separator = ' ')
        {
            this._separator = separator;
            this._stream = stream;
            inputStream = new Queue<string>();
        }
        public string Read
        {
            get
            {
                if (inputStream.Count != 0) return inputStream.Dequeue();
                string[] tmp = _stream.ReadLine().Split(_separator);
                for (int i = 0; i < tmp.Length; ++i)
                    inputStream.Enqueue(tmp[i]);
                return inputStream.Dequeue();
            }
        }
        public string ReadLine { get { return _stream.ReadLine(); } }
        public int ReadInt { get { return int.Parse(Read); } }
        public long ReadLong { get { return long.Parse(Read); } }
        public double ReadDouble { get { return double.Parse(Read); } }
        public string[] ReadStrArray(long N) { var ret = new string[N]; for (long i = 0; i < N; ++i) ret[i] = Read; return ret; }
        public int[] ReadIntArray(long N) { var ret = new int[N]; for (long i = 0; i < N; ++i) ret[i] = ReadInt; return ret; }
        public long[] ReadLongArray(long N) { var ret = new long[N]; for (long i = 0; i < N; ++i) ret[i] = ReadLong; return ret; }
    }
}