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.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		ArrayList<Integer>[] list = new ArrayList[3*n+1];
		for(int i = 1; i <= 3*n; i++){
		    list[i] = new ArrayList<Integer>();
		}
		for(int i = 0; i < m; i++){
		    int from = sc.nextInt();
		    int to = sc.nextInt();
		    list[from].add(to + n);
		    list[from + n].add(to + 2*n);
		    list[from + 2*n].add(to);
		}
		int s = sc.nextInt();
		int t = sc.nextInt();

		// 計算
		int result = -1;
		ArrayDeque<Integer> ad = new ArrayDeque<Integer>();
		int[] visited = new int[3*n+1];
		Arrays.fill(visited, Integer.MAX_VALUE);
		ad.add(s);
		visited[s] = 0;
		while(!ad.isEmpty()){
		    Integer now = ad.poll();
		    for(Integer i : list[now]){
		        int before = visited[i];
		        visited[i] = min(visited[i], visited[now]+1);
		        if(before != visited[i]) ad.add(i);
		        if(i == t) break;
		    }
		}
		if(visited[t] != Integer.MAX_VALUE) result = visited[t]/3;

		// 出力
		System.out.println(result);

	}
}
