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.Scanner;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.ArrayDeque;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		int[][] visit=new int[n][3];
		ArrayList<Integer>[] map=new ArrayList[n];
		for(int i=0; i<n; i++){
			map[i]=new ArrayList<>();
			Arrays.fill(visit[i],Integer.MAX_VALUE);
		}
		for(int i=0; i<m; i++){
			int u=sc.nextInt()-1;
			int v =sc.nextInt()-1;
			map[u].add(v);
		}
		int s =sc.nextInt()-1;
		int t =sc.nextInt()-1;
		ArrayDeque<Integer>  que=new ArrayDeque<>();
		que.add(s);
		int count1=0;
		int count2=0;
		int count3=0;
		while(!que.isEmpty()){
			int j=que.poll();
			if(visit[j][count1%3]>count1){
				visit[j][count1%3]=count1;
				for(int i:map[j]){
					count3++;
					que.add(i);
				}
			}
			if(count2==0){
				count2=count3-1;
				count3=0;
				count1++;
			}else{
				count2--;
			}
		}
		if(visit[t][0]!=Integer.MAX_VALUE){
			System.out.println(visit[t][0]/3);
		}else{
			System.out.println(-1);
		}
	}
}


