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.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Diagnostics;

static class CodeJam
{
    class Node
    {
        public static Node[] nodes;
        public readonly List<int> next1 = new List<int>();
        private List<int> next2;
        private List<int> next3;

        public List<int> Next2()
        {
            if (next2 == null)
            {
                next2 = new List<int>();
                foreach (int i in next1)
                {
                    next2.AddRange(nodes[i].next1);
                }
            }
            return next2;
        }

        public List<int> Next3()
        {
            if (next3 == null)
            {
                next3 = new List<int>();
                foreach (int i in next1)
                {
                    next3.AddRange(nodes[i].Next2());
                }
            }
            return next3;
        }
    }

    static int Solve(int N, int s, int t)
    {
        const int INF = int.MaxValue / 2;
        var d = Enumerable.Repeat(INF, N).ToArray();
        var q = new Queue<int>();
        d[s] = 0;
        q.Enqueue(s);
        while (q.Count > 0)
        {
            int a = q.Dequeue();
            foreach (int b in Node.nodes[a].Next3())
            {
                if (d[a] + 1 < d[b])
                {
                    d[b] = d[a] + 1;
                    q.Enqueue(b);
                }
            }
        }
        return (d[t] < INF) ? d[t] : -1;
    }

    static void Main(string[] args)
    {
        var V1 = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
        int N = V1[0], M = V1[1];
        Node.nodes = new Node[N];
        for (int i = 0; i < N; ++i) { Node.nodes[i] = new Node(); }
        for (int i = 0; i < M; ++i)
        {
            var V2 = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
            var u = V2[0] - 1;
            var v = V2[1] - 1;
            Node.nodes[u].next1.Add(v);
        }
        var V3 = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
        Console.WriteLine(Solve(N, V3[0] - 1, V3[1] - 1));
    }
}