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.*;

public class Main {
	static class Pair {
		int a, b;

		public Pair(int a, int b) {
			this.a = a;
			this.b = b;
		}
		
	}
	static class Graph {
		int N;
		ArrayList<ArrayList<Integer>> node;
		public Graph(int N) {
			this.N = N;
			node = new ArrayList<>();
			for(int i=0; i<N; i++)
				node.add(new ArrayList<Integer>());
		}
		
		public void addEdge(int from, int to) {
			node.get(from).add(to);
		}
		
		public int distance(int s, int t) {
			boolean[] visited = new boolean[N];
			LinkedList<Pair> q = new LinkedList<>();
			q.add(new Pair(s, 0));
			
			while(q.size()>0) {
				Pair p = q.poll();
				if(p.a==t)
					return p.b;
				if(visited[p.a])
					continue;
				for(int to : node.get(p.a)) {
					if(!visited[to])
						q.add(new Pair(to, p.b+1));
				}
				visited[p.a] = true;
			}
			
			return -3;
		}
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		Graph g = new Graph(N*3);
		for(int i=0; i<M; i++) {
			int u = sc.nextInt()-1;
			int v = sc.nextInt()-1;
			g.addEdge(u*3, v*3+1);
			g.addEdge(u*3+1, v*3+2);
			g.addEdge(u*3+2, v*3);
		}
		int S = sc.nextInt()-1;
		int T = sc.nextInt()-1;
		
		System.out.println(g.distance(S*3, T*3)/3);

		sc.close();
	}
}
