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: Runtime Error)

import java.io.*;
import java.lang.reflect.Array;
import java.util.*;

public class Main {
    PrintWriter out = new PrintWriter(System.out);
    long mod = 1000000000+7;
    int n;
    long[] frac, inv;
    public static void main(String[] args) throws Exception {
        Main main = new Main();
//        main.solve();
        System.out.println(main.solve());
    }
    long solve(){
        // it's all about implementation
        Scanner sc = new Scanner(System.in);
        this.n = sc.nextInt();
        int m = sc.nextInt();
        Set<Long> edges = new HashSet<>(m);
        for(int i=0;i<m;i++){
            // u,v < n
            long u = sc.nextLong()-1, v = sc.nextLong()-1;
            edges.add(u*n+v);
        }
        int S = sc.nextInt()-1, T = sc.nextInt()-1;
        ArrayList<Integer>[] graph = GraphFromEdges(edges);
        // use bfs to get new graph
        edges = GetNewEdges(graph);
        graph = GraphFromEdges(edges);
        int step = 0;
        List<Integer> cur = new LinkedList<>(), next = new LinkedList<>();
        cur.add(S);
        boolean[] vis = new boolean[n];
        vis[S] = true;
        while(cur.size()>0){
            for(int now:cur){
                for(int to:graph[now]){
                    if(vis[to]) continue;
                    if(to==T) return step+1;
                    vis[to] = true;
                    next.add(to);
                }
            }
            step += 1;
            cur = next;
            next = new LinkedList<>();
        }
        return -1;
    }
    ArrayList<Integer>[] GraphFromEdges(Set<Long> edges){
        ArrayList<Integer>[] ans = new ArrayList[n];
        for(int i=0;i<n;i++) ans[i] = new ArrayList<>();
        for(long e:edges){
            int u = (int) e/n, v = (int) e%n;
            ans[u].add(v);
        }
        return ans;
    }
    Set<Long> GetNewEdges(ArrayList<Integer>[] graph){
        Set<Long> ans = new HashSet<>(n);
        for(int start=0;start<n;start++){
            HashSet<Integer> cur = new HashSet<>(n/2+1), next = new HashSet<>(n/2+1);
            cur.add(start);
            for(int step=0;step<3&&cur.size()>0;step++){
                for(int now:cur){
                    for(int to:graph[now]) next.add(to);
                }
                cur = next;
                next = new HashSet<>(n/2+1);
            }
            for(int to:cur){
                if(to==start) continue;
                ans.add(start*(long)n+to);
            }
        }
        return ans;
    }
}