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.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int M = sc.nextInt();
        int[] u = new int[M];
        int[] v = new int[M];
        for (int i=0; i<M; i++) {
            u[i] = sc.nextInt();
            v[i] = sc.nextInt();
        }
        int S = sc.nextInt();
        int T = sc.nextInt();

        System.out.println(solve(N, M, u, v, S, T));
    }

    /**
     * そのまま解く．
     * https://atcoder.jp/contests/abc132/submissions/6178008
     * @param N
     * @param M
     * @param u
     * @param v
     * @param S
     * @param T
     * @return
     */
    private static int solve(int N, int M, int[] u, int[] v, int S, int T) {
        List<List<Integer>> graph = new ArrayList();
        for (int i=0; i<N; i++) {
            graph.add(new ArrayList());
        }
        for (int i=0; i<M; i++) {
            graph.get(u[i]-1).add(v[i]-1);
        }
        // graph.forEach(System.err::println);


        List<Integer> next = new ArrayList();
        next.add(S-1);
        List<Integer> next2 = new ArrayList();

        boolean[][] reached = new boolean[3][N];

        // i 回目のけんけんぱで移動できる範囲を列挙する
        for (int i=1; !next.isEmpty(); i++) {
            for (int k=0; k<3; k++) {
                for (int s: next) {
                    if(!reached[k][s]) {
                        next2.addAll(graph.get(s));
                        reached[k][s] = true;
                    }
                }

                next.clear();
                next.addAll(next2);
                next2.clear();
            }

            for (Integer n: next) {
                if (n == T-1) return i;

                next2.add(n);
            }

            next.clear();
            next.addAll(next2);
            next2.clear();
        }

        return -1;
    }
}
