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.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {

    long MOD = (long) 1e9 + 7;

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

        Edges[] g = new Edges[n];

        for (int i = 0; i < n; i++) g[i] = new Edges();

        for (int i = 0; i < m; i++) {
            int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
            g[u].add(v);
        }

        int s = sc.nextInt() - 1, t = sc.nextInt() - 1;
        System.out.println(dijkstra(g, s, t, n));
    }

    int INF = 1 << 28;
    int dijkstra(Edges[] g, int s, int t, int n) {
        int[][] d = new int[n][3];
        for (int i = 0; i < n; i++) Arrays.fill(d[i], INF);
        PriorityQueue<Node> q = new PriorityQueue<>();

        q.add(new Node(s, 2));
        d[s][2] = 2;
        while (!q.isEmpty()) {
            Node cur = q.remove();
//            debug(cur.idx, cur.c);
            if (d[cur.idx][cur.c % 3] != cur.c) continue;
            if (cur.idx == t && cur.c % 3 == 2) return cur.c / 3;

            for (int next: g[cur.idx]) if (d[next][(cur.c + 1) % 3] > cur.c + 1) {
                q.add(new Node(next, cur.c + 1));
                d[next][(cur.c + 1) % 3] = cur.c + 1;
            }

        }
        return -1;
    }

    class Node implements Comparable<Node> {
        int idx, c;

        Node(int idx, int c) {
            this.idx = idx;
            this.c = c;
        }

        @Override
        public int compareTo(Node o) {
            return c - o.c;
        }
    }

    class Edges extends ArrayList<Integer> {}

    class Edge {
        int t, c;
        Edge(int t, int c) {
            this.t = t;
            this.c = c;
        }
    }

    static void debug(Object...os) {
        System.err.println(Arrays.deepToString(os));
    }

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