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: Accepted)

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 readonly List<Node> next = new List<Node>();
        public int d = int.MaxValue / 2;
    }

    static int Solve(int N, int M, int[][] edges, int s, int t)
    {
        const int INF = int.MaxValue / 2;
        var nodes = new Node[N][];
        for (int i = 0; i < N; ++i)
        {
            nodes[i] = new Node[3];
            for (int j = 0; j < 3; ++j) { nodes[i][j] = new Node(); }
        }
        foreach (var edge in edges)
        {
            int u = edge[0];
            int v = edge[1];
            nodes[u][0].next.Add(nodes[v][1]);
            nodes[u][1].next.Add(nodes[v][2]);
            nodes[u][2].next.Add(nodes[v][0]);
        }
        var q = new Queue<Node>();
        nodes[s][0].d = 0;
        q.Enqueue(nodes[s][0]);
        while (q.Count > 0)
        {
            Node a = q.Dequeue();
            foreach (Node b in a.next)
            {
                if (a.d + 1 < b.d)
                {
                    b.d = a.d + 1;
                    q.Enqueue(b);
                }
            }
        }
        return (nodes[t][0].d < INF && nodes[t][0].d % 3 == 0) ? nodes[t][0].d / 3 : -1;
    }

    static void Main(string[] args)
    {
        var V1 = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
        int N = V1[0], M = V1[1];
        var edges = new int[M][];
        for (int i = 0; i < M; ++i)
        {
            edges[i] = Array.ConvertAll(Console.ReadLine().Split(), x => int.Parse(x) - 1);
        }
        var V3 = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
        Console.WriteLine(Solve(N, M, edges, V3[0] - 1, V3[1] - 1));
    }
}