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;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Numerics;
using System.Threading.Tasks;
using static System.Console;
using static System.Math;
using static Func;
class Z { static void Main() => new K(); }
class K
{
	int F => int.Parse(Str);
	long FL => long.Parse(Str);
	int[] G => Strs.Select(int.Parse).ToArray();
	uint[] GU => Strs.Select(uint.Parse).ToArray();
	decimal[] GD => Strs.Select(decimal.Parse).ToArray();
	long[] GL => Strs.Select(long.Parse).ToArray();
	string Str => ReadLine();
	string[] Strs => Str.Split(new char[] { ' ', '\t' }, StringSplitOptions.RemoveEmptyEntries);
	static ulong One(int k) => 1ul << k;
	static ulong All(int k) => One(k) - 1;
	static bool Has(ulong s, int k) => (s & One(k)) != 0;
	const int MOD = 1000000007;
	public K()
	{
		SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
		Solve();
		Out.Flush();
	}
	void Solve()
	{
		var I = G;
		int N = I[0], M = I[1];
		var g = MakeArray(N, _ => new List<int>());
		for (var i = 0; i < M; i++)
		{
			I = G;
			g[I[0] - 1].Add(I[1] - 1);
		}
		I = G;
		int S = I[0] - 1, T = I[1] - 1;
		var done = new bool[N, 3];
		var q = new Queue<Pair>();
		q.Enqueue(new Pair(S, 0));
		while (q.Count > 0)
		{
			var p = q.Dequeue();
			if (done[p.V, p.D % 3]) continue;
			done[p.V, p.D % 3] = true;
			if (p.V == T && p.D % 3 == 0) { WriteLine(p.D / 3); return; }
			foreach (var u in g[p.V]) if (!done[u, (p.D + 1) % 3]) q.Enqueue(new Pair(u, p.D + 1));
		}
		WriteLine(-1);
	}
}
struct Pair
{
	public readonly int V, D;
	public Pair(int v, int d) { V = v; D = d; }
}
static class Func
{
	public static T[] ConstantArray<T>(int n, T val) { var a = new T[n]; for (var i = 0; i < n; i++) a[i] = val; return a; }
	public static T[] MakeArray<T>(int n, Func<int, T> f) { var a = new T[n]; for (var i = 0; i < n; i++) a[i] = f(i); return a; }
	public static T[] Recurse<T>(int n, T a0, Func<int, T, T> f) { var a = new T[n]; a[0] = a0; for (var i = 1; i < n; i++) a[i] = f(i, a[i - 1]); return a; }
	public static T[] Accumulate<T>(IList<T> a, T zero, Func<T, T, T> f) { var s = new T[a.Count + 1]; s[0] = zero; for (var i = 0; i < a.Count; i++) s[i + 1] = f(s[i], a[i]); return s; }
	public static long GCD(long a, long b)
	{
		var n = (ulong)Math.Abs(a); var m = (ulong)Math.Abs(b);
		if (n == 0) return (long)m; if (m == 0) return (long)n;
		int zm = 0, zn = 0;
		while ((n & 1) == 0) { n >>= 1; zn++; }
		while ((m & 1) == 0) { m >>= 1; zm++; }
		while (m != n)
		{
			if (m > n) { m -= n; while ((m & 1) == 0) m >>= 1; }
			else { n -= m; while ((n & 1) == 0) n >>= 1; }
		}
		return (long)n << Math.Min(zm, zn);
	}
	public static long Inverse(long a, long mod)
	{
		if (a < 0) { a %= mod; if (a < 0) a += mod; }
		var t = SolveLinear(a, mod);
		return t.Item1 > 0 ? t.Item1 : t.Item1 + mod;
	}
	public static long Pow(long a, long b, long mod)
	{
		var p = 1L;
		var x = a;
		while (b > 0)
		{
			if ((b & 1) == 1) p = (p * x) % mod;
			b >>= 1;
			x = (x * x) % mod;
		}
		return p;
	}
	public static Tuple<long, long> SolveLinear(long n, long m)
	{
		if (n < 0) { var p = SolveLinear(-n, m); return p == null ? p : new Tuple<long, long>(-p.Item1, p.Item2); }
		if (m < 0) { var p = SolveLinear(n, -m); return p == null ? p : new Tuple<long, long>(p.Item1, -p.Item2); }
		if (n < m) { var p = SolveLinear(m, n); return p == null ? p : new Tuple<long, long>(p.Item2, p.Item1); }
		long a = 1, b = 0, c = 0, d = 1;
		while (m > 0)
		{
			var r = n % m;
			var q = n / m;
			n = m;
			m = r;
			var tmp = a;
			a = -a * q + b;
			b = tmp;
			tmp = c;
			c = -c * q + d;
			d = tmp;
		}
		return n != 1 ? null : new Tuple<long, long>(d, b);
	}
}