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.Collections.Generic;
using System.Linq;
using static Input;

public class Prog
{
    private const int INF = 1000000007;
    public struct Pair<T> { public T x, y; public Pair(T x, T y) { this.x = x; this.y = y; } }
    public static void Solve()
    {
        int N = NextInt();
        int M = NextInt();
        Node[] nodes = new Node[N + 1];
        Node[] nx = new Node[N + 1];
        for (int i = 1; i <= N; i++) nodes[i] = new Node(i);
        for (int i = 1; i <= N; i++) nx[i] = new Node(i);
        for (int i = 0; i < M; i++)
        {
            int um = NextInt();
            int vm = NextInt();
            nodes[um].relate.Add(new Edge(vm));
        }

        int S = NextInt();
        int T = NextInt();

        Queue<int> Q = new Queue<int>();
        Q.Enqueue(S);

        Dictionary<int, Dictionary<int, int>> d = new Dictionary<int, Dictionary<int, int>>();


        while (Q.Count() != 0)
        {
            int n = Q.Dequeue();
            if (!d.ContainsKey(n)) d[n] = new Dictionary<int, int>();
            foreach (var e0 in nodes[n].relate)
                //1経路目
                foreach (var e1 in nodes[e0.to].relate)
                    //2経路目
                    foreach (var e2 in nodes[e1.to].relate)
                        //3経路目
                        if (!d[n].ContainsKey(e2.to) && n != e2.to) { d[n][e2.to] = 0; Q.Enqueue(e2.to); }



        }

        //出来ているグラフで最短経路
        foreach (var d0 in d)
            foreach (var d1 in d[d0.Key])
            {
                nx[d0.Key].relate.Add(new Edge(d1.Key));
                Console.WriteLine(d0.Key + " " + d1.Key);
            }

        //ダイクストラ
        Queue<int> PQ = new Queue<int>();

        PQ.Enqueue(S);
        nx[S].cost = 0;

        while (PQ.Count() != 0)
        {
            int n = PQ.Dequeue();
            if (n == T) { Console.WriteLine(nx[n].cost); return; }
            nx[n].visit = true;
            foreach (var e in nx[n].relate)
            {
                if (!nx[e.to].visit)
                {
                    nx[e.to].cost = Math.Min(nx[e.to].cost, nx[n].cost + 1);
                    PQ.Enqueue(e.to);
                }
            }
        }

        //for (int i = 1; i <= N; i++) Console.WriteLine(nx[i].cost);

        Console.WriteLine(-1);

    }
    public static void Main()
    {
        var sw = new System.IO.StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
        Console.SetOut(sw);
        Solve();
        Console.Out.Flush();
    }
}

public class PriorityQueue<Tkey, Tvalue> where Tkey : IComparable
{
    private List<Tkey> keys = new List<Tkey>();
    private List<Tvalue> values = new List<Tvalue>();
    private int size = 0;
    private static bool Descending;
    private static int Compare(Tkey a, Tkey b) { return (Descending ? a.CompareTo(b) : b.CompareTo(a)); }

    public PriorityQueue(bool b = true) { Clear(); Descending = b; }
    public int Count() { return size; }
    public void Clear() { keys = new List<Tkey>(); values = new List<Tvalue>(); size = 0; }
    public void Add(Tkey e, Tvalue v)
    {
        keys.Add(e);
        values.Add(v);
        int x = size;
        size += 1;
        while (x != 0)
        {
            int par = (x % 2 == 0 ? x - 2 : x - 1) / 2;
            if (Compare(keys[par], e) <= 0) break;
            Tkey a = keys[x]; keys[x] = keys[par]; keys[par] = a;
            Tvalue b = values[x]; values[x] = values[par]; values[par] = b;
            x = par;
        }
    }
    public void Poll()
    {
        Tkey k = keys[size - 1];
        Tvalue v = values[size - 1];
        int x = 0;
        size--;
        if (size == 0) { Clear(); return; }
        keys.RemoveAt(size);
        values.RemoveAt(size);
        while (true)
        {
            int l = x * 2 + 1;
            int r = x * 2 + 2;
            if (size - 1 < l) break;
            int y = 0;
            if (size - 1 == l) { y = l; }
            else if (Compare(keys[l], keys[r]) == 1) { y = r; }
            else { y = l; }
            if (Compare(keys[y], k) > 0) break;
            keys[x] = keys[y];
            values[x] = values[y];
            x = y;
        }
        keys[x] = k;
        values[x] = v;
    }
    public Tvalue Take() { Tvalue ret = values[0]; Poll(); return ret; }
    public Tvalue Peek() { return values[0]; }
    public void Debug()
    {
        Console.Write("key   : ");
        foreach (var k in keys) Console.Write(k + " ");
        Console.WriteLine();
        Console.Write("value : ");
        foreach (var k in values) Console.Write(k + " ");
        Console.WriteLine();
    }
}


public class Node
{
    public List<Edge> relate = new List<Edge>();
    public int name = 0;
    public bool visit = false;
    public int cost = 1000000007;
    public Node(int name) { this.name = name; }
}
public class Edge
{
    public int to;
    public Edge(int to) { this.to = to; }
}

public class Input
{
    private static Queue<string> q = new Queue<string>();
    private static void Confirm() { if (q.Count == 0) foreach (var s in Console.ReadLine().Split(' ')) q.Enqueue(s); }
    public static int NextInt() { Confirm(); return int.Parse(q.Dequeue()); }
    public static long NextLong() { Confirm(); return long.Parse(q.Dequeue()); }
    public static char NextChar() { Confirm(); return char.Parse(q.Dequeue()); }
    public static string NextString() { Confirm(); return q.Dequeue(); }
    public static double NextDouble() { Confirm(); return double.Parse(q.Dequeue()); }
    public static int[] LineInt() { return Console.ReadLine().Split(' ').Select(int.Parse).ToArray(); }
    public static long[] LineLong() { return Console.ReadLine().Split(' ').Select(long.Parse).ToArray(); }
    public static string[] LineString() { return Console.ReadLine().Split(' ').ToArray(); }
    public static double[] LineDouble() { return Console.ReadLine().Split(' ').Select(double.Parse).ToArray(); }
}