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 {
	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	class P {
		int v;
		int l;
		P(int v, int l){
			this.v=v;
			this.l=l;
		}
	}

	public void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();

		ArrayList<ArrayList<Integer>> g = new ArrayList<>();
		for(int i=0; i<n; i++) g.add(new ArrayList<>());
		for(int i=0; i<m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			a--;
			b--;
			g.get(a).add(b);
		}
		int s = sc.nextInt();
		int t = sc.nextInt();
		s--;
		t--;
		int dist[][] = new int[n][3];
		int INF = 1001001001;
		for(int i=0; i<n; i++) {
			for(int j=0; j<3; j++) {
				dist[i][j] = INF;
			}
		}
		ArrayDeque<P> q = new ArrayDeque<>();
		dist[s][0] = 0;
		q.add(new P(s,0));
		while(! q.isEmpty()) {
			P p = q.pop();
			int v = p.v;
			int l = p.l;
			int nl = (l+1)%3;
			for (int nv : g.get(v)) {
				if(dist[nv][nl] != INF) continue;
				dist[nv][nl] = dist[v][l] + 1;
				q.add(new P(nv, nl));
			}
		}
		if(dist[t][0] == INF) {
			System.out.println(-1);
			sc.close();
			return;
		}

		int ans = dist[t][0] / 3;
		System.out.println(ans);
		sc.close();
	}

}
