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.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        Node[] nodes = new Node[n+1];
        for(int i = 1; i <= n; i++) {
            nodes[i] = new Node(i);
        }
        for(int i = 0; i < m; i++) {
            int src = sc.nextInt();
            int dest = sc.nextInt();
            nodes[src].nexts.add(dest);
        }
        int s = sc.nextInt();
        int t = sc.nextInt();
        Node start = nodes[s];
        start.minCount = 0;
        ArrayDeque<Node> queue = new ArrayDeque<>();
        queue.addLast(start);
        while(!queue.isEmpty()) {
            Node curr = queue.removeFirst();
            if(curr.num == t) {
                System.out.println(curr.minCount);
                return;
            }
            int prev = curr.minCount;
            for(int next1: curr.nexts) {
                for(int next2: nodes[next1].nexts) {
                    for(int next3: nodes[next2].nexts) {
                        if(nodes[next3].minCount != -1) continue;
                        nodes[next3].minCount = prev + 1;
                        queue.addLast(nodes[next3]);
                    }
                }
            }
        }
        System.out.println(-1);
    }
    static class Node{
        int num;
        int minCount;
        ArrayList<Integer> nexts;
        Node(int num) {
            this.num = num;
            minCount = -1;
            nexts = new ArrayList<>();
        }
    }
}