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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Globalization;
using System.IO;
using System.Linq;

internal partial class Solver {
    public void Run() {
        var N = ni();
        var M = ni();
        var adj = Enumerable.Range(0, N).Select(_ => new List<int>()).ToArray();
        for (int i = 0; i < M; i++) {
            var u = ni() - 1;
            var v = ni() - 1;
            adj[u].Add(v);
        }

        var S = ni();
        var T = ni();
        S--; T--;

        var kenkenpaAdj = Enumerable.Range(0, N).Select(_ => new List<int>()).ToArray();
        for (int i = 0; i < N; i++) {
            var qu = new Queue<int>();
            qu.Enqueue(i);
            qu.Enqueue(-1);
            qu.Enqueue(0);
            while (qu.Count > 0) {
                var pos = qu.Dequeue();
                var parent = qu.Dequeue();
                var len = qu.Dequeue();
                if (len == 3) {
                    if (i != pos) {
                        kenkenpaAdj[i].Add(pos);
                    }
                    continue;
                }
                foreach (var next in adj[pos]) {
                    qu.Enqueue(next);
                    qu.Enqueue(pos);
                    qu.Enqueue(len + 1);
                }
            }
        }

        cout.WriteLine(Bfs(kenkenpaAdj, S, T));
    }

    private static int Bfs(List<int>[] adj, int s, int t) {
        var n = adj.Length;
        var qu = new Queue<int>();
        qu.Enqueue(s);
        qu.Enqueue(0);
        var visited = new bool[n];
        while (qu.Count > 0) {
            var pos = qu.Dequeue();
            var len = qu.Dequeue();
            if (visited[pos]) continue;
            visited[pos] = true;
            if (pos == t) return len;
            foreach (var next in adj[pos]) {
                qu.Enqueue(next);
                qu.Enqueue(len + 1);
            }
        }
        return -1;
    }
}

// PREWRITEN CODE BEGINS FROM HERE
internal partial class Solver : Scanner {
    public static void Main(string[] args) {
#if LOCAL
        new Solver(Console.In, Console.Out).Run();
#else
        Console.SetOut(new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });
        new Solver(Console.In, Console.Out).Run();
        Console.Out.Flush();
#endif
    }

    public void RunMulti() {
        while (HasNext()) {
            Run();
        }
    }

#pragma warning disable IDE0052
    private readonly TextReader cin;
    private readonly TextWriter cout;
#pragma warning restore IDE0052

    public Solver(TextReader reader, TextWriter writer)
        : base(reader) {
        cin = reader;
        cout = writer;
    }
    public Solver(string input, TextWriter writer)
        : this(new StringReader(input), writer) {
    }

#pragma warning disable IDE1006
#pragma warning disable IDE0051
    private int ni() { return NextInt(); }
    private int[] ni(int n) { return NextIntArray(n); }
    private long nl() { return NextLong(); }
    private long[] nl(int n) { return NextLongArray(n); }
    private double nd() { return NextDouble(); }
    private double[] nd(int n) { return NextDoubleArray(n); }
    private string ns() { return Next(); }
    private string[] ns(int n) { return NextArray(n); }
#pragma warning restore IDE1006
#pragma warning restore IDE0051
}

public static class LinqPadExtension {
    public static T Dump<T>(this T obj) {
#if LOCAL
        return LINQPad.Extensions.Dump(obj);
#else
        return obj;
#endif
    }
}
public class Scanner {
    private readonly TextReader Reader;
    private readonly Queue<string> TokenQueue = new Queue<string>();
    private readonly CultureInfo ci = CultureInfo.InvariantCulture;

    public Scanner()
        : this(Console.In) {
    }

    public Scanner(TextReader reader) {
        Reader = reader;
    }

    public int NextInt() { return int.Parse(Next(), ci); }
    public long NextLong() { return long.Parse(Next(), ci); }
    public double NextDouble() { return double.Parse(Next(), ci); }
    public string[] NextArray(int size) {
        var array = new string[size];
        for (var i = 0; i < size; i++) {
            array[i] = Next();
        }

        return array;
    }
    public int[] NextIntArray(int size) {
        var array = new int[size];
        for (var i = 0; i < size; i++) {
            array[i] = NextInt();
        }

        return array;
    }

    public long[] NextLongArray(int size) {
        var array = new long[size];
        for (var i = 0; i < size; i++) {
            array[i] = NextLong();
        }

        return array;
    }

    public double[] NextDoubleArray(int size) {
        var array = new double[size];
        for (var i = 0; i < size; i++) {
            array[i] = NextDouble();
        }

        return array;
    }

    public string Next() {
        if (TokenQueue.Count == 0) {
            if (!StockTokens()) {
                throw new InvalidOperationException();
            }
        }
        return TokenQueue.Dequeue();
    }

    public bool HasNext() {
        if (TokenQueue.Count > 0) {
            return true;
        }

        return StockTokens();
    }

    private static readonly char[] _separator = new[] { ' ' };
    private bool StockTokens() {
        while (true) {
            var line = Reader.ReadLine();
            if (line == null) {
                return false;
            }

            var tokens = line.Split(_separator, StringSplitOptions.RemoveEmptyEntries);
            if (tokens.Length == 0) {
                continue;
            }

            foreach (var token in tokens) {
                TokenQueue.Enqueue(token);
            }

            return true;
        }
    }
}