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 System.Console;
using static System.Math;

class Program
{
    static void Main()
    {
        var NM = ReadLine().Split().Select(int.Parse).ToArray();
        var N = NM[0];
        var M = NM[1];
        var g = new List<int>[N];
        for (int i = 0; i < g.Length; i++)
        {
            g[i] = new List<int>();
        }
        for (int i = 0; i < M; i++)
        {
            var uv = ReadLine().Split(' ');
            g[int.Parse(uv[0]) - 1].Add(int.Parse(uv[1]) - 1);
        }
        var ST = ReadLine().Split().Select(int.Parse).ToArray();
        var S = ST[0] - 1;
        var T = ST[1] - 1;

        var vis = new int[N];
        var q = new Queue<int>();
        q.Enqueue(S);
        while (q.Count != 0)
        {
            var v = q.Dequeue();
            var d = vis[v];
            if (d != 0) continue;
            foreach (var u1 in g[v])
                foreach (var u2 in g[u1])
                    foreach (var u3 in g[u2])
                    {
                        if (vis[u3] == 0)
                        {
                            vis[u3] = d + 1;
                            q.Enqueue(u3);
                        }
                    }
        }

        WriteLine(vis[T] == 0 ? -1 : vis[T]);
    }
}