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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

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

    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        int M = scan.nextInt();
        HashMap<Integer, List<Integer>> graph = new HashMap<Integer, List<Integer>>();
        for (int i = 0; i < M; i++) {
            int f = scan.nextInt()-1;
            int t = scan.nextInt()-1;
            if(!graph.containsKey(f)) {
                graph.put(f, new ArrayList<>());
            }
            graph.get(f).add(t+1000000);
            if(!graph.containsKey(f+1000000)) {
                graph.put(f+1000000, new ArrayList<>());
            }
            graph.get(f+1000000).add(t+2000000);
            if(!graph.containsKey(f+2000000)) {
                graph.put(f+2000000, new ArrayList<>());
            }
            graph.get(f+2000000).add(t);
        }
        int S = scan.nextInt() - 1;
        int T = scan.nextInt() - 1;
        int[][] d_S = dijkstra(graph, N, S);
        if (d_S[T][0] == 100000000) {
            System.out.println(-1);
        } else {
            System.out.println(d_S[T][0]/3);
        }
    }
    public int[][] dijkstra(HashMap<Integer, List<Integer>> graph, int N, int S) {
        int[][] d = new int[N][3];
        boolean[][] used = new boolean[N][3];
        for (int i = 0; i < N; i++) {
            Arrays.fill(d[i], 100000000);
            Arrays.fill(used[i], false);
        }
        d[S][0] = 0;
        Queue<Integer> q = new LinkedList<>();
        q.add(S);
        while(!q.isEmpty()) {
            int p = q.poll();
            int x = p % 1000000;
            int y = p / 1000000;
            if (used[x][y]) {
                continue;
            }
            used[x][y] = true;
            List<Integer> nexts = graph.get(p);
            if (nexts == null) {
                continue;
            }
            for (int n : nexts) {
                int nx = n % 1000000;
                int ny = n / 1000000;
                d[nx][ny] = Math.min(d[nx][ny], d[x][y] + 1);
                q.add(n);
            }
        }
        return d;
    }
}
