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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.io.*;

public class Main {
	static int n;
	static ArrayList<Integer>[] graph;
	static int[] place;
	static int[] place2;
	static int[] place3;
	static int t;
	public static void main (String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] first = br.readLine().split(" ", 2);
		n = Integer.parseInt(first[0]);
		int m = Integer.parseInt(first[1]);
		graph = new ArrayList[n + 1];
		place = new int[n + 1];
		place2 = new int[n + 1];
		place3 = new int[n + 1];
		for (int i = 1; i <= n; i++) {
			graph[i] = new ArrayList<Integer>();
			place[i] = -1;
			place2[i] = -1;
			place3[i] = -1;
		}
		for (int i = 0; i < m; i++) {
			String[] line = br.readLine().split(" ", 2);
			int a = Integer.parseInt(line[0]);
			int b = Integer.parseInt(line[1]);
			graph[a].add(b);
		}
		String[] last = br.readLine().split(" ", 2);
		int s = Integer.parseInt(last[0]);
		t = Integer.parseInt(last[1]);
		next(s, 0, 0);
		System.out.println(place[t]);
		
	}
	
	static void next(int idx, int num, int cnt) {
		if (cnt == 0) {
			if (place[idx] != -1 && place[idx] <= num) {
				return;
			}
			place[idx] = num;
			if (idx == t) {
				return;
			}
			num++;
			cnt = 3;
		} else if (cnt == 1) {
			if (place2[idx] != -1 && place2[idx] <= num) {
				return;
			}
			place2[idx] = num;
		} else if (cnt == 2) {
			if (place3[idx] != -1 && place3[idx] <= num) {
				return;
			}
			place3[idx] = num;
		}
		for (int x : graph[idx]) {
			next(x, num, cnt - 1);
		}
	}
}