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 {
    private static final int INF = 1001001001;
    private static int dist[][] = new int[100005][3];

    public static void main(String... args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        List<Integer> to[] = new ArrayList[n];
        for (int i=0;i<n;i++) {
            to[i] = new ArrayList<>();
        }

        for (int i = 0; i < m; i++) {
            int u = sc.nextInt();
            int v = sc.nextInt();
            u--;
            v--;
            to[u].add(v);
        }

        int s = sc.nextInt();
        int t = sc.nextInt();
        s--;
        t--;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 3; j++) {
                dist[i][j] = INF;
            }
        }

        // BSF
        Queue<Pair<Integer, Integer>> q = new LinkedList<>();

        q.add(new Pair<>(s, 0));
        dist[s][0] = 0;
        while (!q.isEmpty()) {
            int v = q.peek().first;
            int l = q.peek().second;
            q.poll();
            for (int u : to[v]) {
                int nl = (l + 1) % 3;
                if (dist[u][nl] != INF) continue;
                dist[u][nl] = dist[v][l] + 1;
                q.add(new Pair<>(u, nl));
            }
        }
        int ans = dist[t][0];
        if (ans == INF) ans = -1;
        else ans /= 3;
        System.out.println(ans);
    }

    static class Pair<F, S> {
        public final F first;
        public final S second;

        Pair(F first, S second) {
            this.first = first;
            this.second = second;
        }
    }
}