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;

namespace EHopscotchAddict
{
    class Program
    {
        static void Main(string[] args)
        {
            // 入力の取得
            var inputNM = Console.ReadLine().Split().ToArray();
            var n = int.Parse(inputNM[0]);
            var m = long.Parse(inputNM[1]);

            var nodes = new List<int>[n + 1];
            for (var i = 0; i < n + 1; i++) nodes[i] = new List<int>();
            for (var i = 0; i < m; i++)
            {
                var inputUV = Console.ReadLine().Split().Select(int.Parse).ToArray();
                int u = inputUV[0], v = inputUV[1];
                nodes[u].Add(v);
            }

            var inputST = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int s = inputST[0], t = inputST[1];

            var distances = new int[n + 1, 3];
            var queue = new Queue<Tuple<int, int>>();

            for (var i = 0; i < n + 1; i++)
            {
                distances[i, 0] = -1;
                distances[i, 1] = -1;
                distances[i, 2] = -1;
            }
            distances[s, 0] = 0;
            queue.Enqueue(Tuple.Create(s, 0));

            while (queue.Any())
            {
                var path = queue.Dequeue();
                int node = path.Item1, state = path.Item2;

                var nextState = (state + 1) % 3;
                foreach (var dest in nodes[node])
                {
                    if (distances[dest, nextState] != -1) continue;
                    distances[dest, nextState] = distances[node, state] + 1;
                    queue.Enqueue(Tuple.Create(dest, nextState));
                }
            }

            var goalDistance = distances[t, 0];
            var answer = goalDistance < 0 ? -1 : goalDistance / 3;

            // 解答の出力
            Console.WriteLine(answer);
        }
    }
}