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.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;

public class Main {
	static class Node{
		List<Integer> adj;
		public Node() {
			adj = new ArrayList<Integer>();
		}
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		Node vs[] = new Node[N * 3];
		for(int i = 0 ; i < N ; ++i){
			vs[i] = new Node();
			vs[i + N] = new Node();
			vs[i + 2 * N] = new Node();
		}
		for(int i = 0 ; i < M ; ++i){
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			vs[a].adj.add(b + N);
			vs[a + N].adj.add(b + 2 * N);
			vs[a + 2 * N].adj.add(b);
		}
		int S = sc.nextInt() - 1;
		int T = sc.nextInt() - 1;
		int dist[] = new int[vs.length];
		Arrays.fill(dist, -1);
		dist[S] = 0;
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(S);
		while(!q.isEmpty()){
			int cur = q.poll();
			int cd = dist[cur];
			for(int a : vs[cur].adj){
				if(dist[a] < 0){
					dist[a] = cd + 1;
					q.add(a);
				}
			}
		}
		if(dist[T] >= 0){
			System.out.println(dist[T] / 3);			
		}else{
			System.out.println(dist[T]);			
		}
	}
}
