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.*;

class Main{
	public static void main(String[] $){
		System.err.println();
		Scanner s=new Scanner(System.in);

		ArrayList<ArrayList<Integer>> graph=new ArrayList<>();
		int n=s.nextInt(),m=s.nextInt();

		for(int i=0;i<n;++i)
			graph.add(new ArrayList<>(0));

		for(int i=0;i<m;++i){
			int f=s.nextInt()-1;
			int t=s.nextInt()-1;
			graph.get(f).add(t);
		}
		//graph.forEach(System.err::println);

		int f=s.nextInt()-1,t=s.nextInt()-1;

		boolean[][] used=new boolean[3][n];

		ArrayList<Integer> next=new ArrayList<>();
		ArrayList<Integer> nextnext=new ArrayList<>();
		next.add(f);
		for(int i=1;!next.isEmpty();++i){
			for(int kenken=0;kenken<3;++kenken){
				for(int p:next){
					for(int q:graph.get(p)){
						if(!used[kenken][q]){
							used[kenken][q]=true;
							nextnext.add(q);
						}
					}
				}
				next.clear();
				next.addAll(nextnext);
				nextnext.clear();
			}

			if(used[2][t]){
				System.out.println(i);
				return;
			}
		}
		System.out.println(-1);
	}
}