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 System.Text.RegularExpressions;
using System.Threading.Tasks;

class Program
{
    int n, m;
    List<List<int>> g;
    int s, t;
    int[] d;
    bool[] reachable;
    bool[] visit;
    int MAX = 1 << 29;
    void Run()
    {
        var sc = new Scanner();
        n = sc.NextInt();
        m = sc.NextInt();
        g = new List<List<int>>();
        for (int i = 0; i < n; i++) g.Add(new List<int>());
        for (int i = 0; i < m; i++)
        {
            int a = sc.NextInt() - 1;
            int b = sc.NextInt() - 1;
            g[a].Add(b);
            //g[b].Add(a);
        }
        s = sc.NextInt() - 1;
        t = sc.NextInt() - 1;

        d = new int[n];
        reachable = new bool[n];
        visit = new bool[n];
        for (int i = 0; i < n; i++)
            d[i] = MAX;

        var q = new Queue<int>();
        q.Enqueue(s);
        while (q.Count > 0)
        {
            int v = q.Dequeue();
            if (reachable[v]) continue;
            reachable[v] = true;
            foreach (int u in g[v])
                q.Enqueue(u);
        }

        if (!reachable[t])
        {
            Console.WriteLine(-1);
            return;
        }

        bool changed = true;
        Dfs(s, 0, 1);
        while (changed)
        {
            changed = false;
            for (int i = 0; i < n; i++)
            {
                if (d[i] != MAX) changed |= Dfs(i, 0, d[i] + 1);
            }
        }
        Console.WriteLine(d[t] == MAX ? -1 : d[t]);
    }

    bool Dfs(int v, int dep, int c)
    {
        if (visit[v]) return false;
        visit[v] = true;
        if (dep == 3)
        {
            visit[v] = false;
            if (d[v] <= c)
            {
                return false;
            }
            d[v] = c;
            return true;
        }
        bool res = false;
        foreach (int u in g[v])
            res |= Dfs(u, dep + 1, c);
        visit[v] = false;
        return res;
    }

    static void Main(string[] args)
    {
        new Program().Run();
    }

    static long Gcd(long a, long b)

    {
        if (a < b) return Gcd(b, a);
        if (b == 0) return a;
        return Gcd(b, a % b);
    }

    static long Lcm(long a, long b)
    {
        long d = Gcd(a, b);
        return a / d * b;
    }
}

class Scanner
{
    int index = -1;
    string[] buf;

    public Scanner()
    {
    }

    public int NextInt()
    {
        CheckNext();
        return int.Parse(buf[index++]);
    }

    public long NextLong()
    {
        CheckNext();
        return long.Parse(buf[index++]);
    }

    public double NextDouble()
    {
        CheckNext();
        return double.Parse(buf[index++]);
    }

    public string NextString()
    {
        CheckNext();
        return buf[index++];
    }

    private void CheckNext()
    {
        if (index >= 0 && index < buf.Length) return;

        buf = Console.ReadLine().Split(' ');
        index = 0;
    }
}