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.awt.Point;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int m = Integer.parseInt(sc.next());
        List<List<Integer>> edge = new ArrayList<>();
        for(int i = 0; i < n; i++) {
            edge.add(new ArrayList<>());
        }
        for(int i = 0; i < m; i++) {
            int u = Integer.parseInt(sc.next()) - 1;
            int v = Integer.parseInt(sc.next()) - 1;
            edge.get(u).add(v);
        }
        int[][] d = new int[n][3];
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < 3; j++) {
                d[i][j] = -1;
            }
        }
        int s = Integer.parseInt(sc.next()) - 1;
        int t = Integer.parseInt(sc.next()) - 1;
        d[s][0] = 0;
        Queue<Point> que = new ArrayDeque<>();
        que.add(new Point(s, 0));
        boolean flag = true;
        loop : while(!que.isEmpty()) {
            Point p = que.poll();
            List<Integer> to = edge.get(p.x);
            for(int x : to) {
                if(x == t && (p.y + 1) % 3 == 0) {
                    System.out.println((p.y + 1) / 3);
                    flag = false;
                    break loop;
                }
                if(d[x][(p.y + 1) % 3] == -1) {
                    d[x][(p.y + 1) % 3] = p.y + 1;
                    que.add(new Point(x, p.y + 1));
                }
            }
        }
        if(flag) System.out.println(-1);
    }
}