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 {
  
  public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();

    List<List<Integer>> nodes = new ArrayList<>();
    for (int i = 0; i < 3*(N+1); i++) {
      nodes.add(new ArrayList<>());
    }

    int[] minDist = new int[3*(N+1)];
    for (int i = 0; i < minDist.length; i++) {
      minDist[i] = -1;
    }

    for (int i = 0; i < M; i++) {
      int u = sc.nextInt();
      int v = sc.nextInt();
      nodes.get(3*u).add(3*v+1);
      nodes.get(3*u+1).add(3*v+2);
      nodes.get(3*u+2).add(3*v);
    }

    int S = sc.nextInt();
    int T = sc.nextInt();
    
    minDist[3*S] = 0;

    // BSF
    LinkedList<Integer> queue = new LinkedList<>();
    queue.add(3*S);
    while (!queue.isEmpty()) {
      int n = queue.remove();
      if (n == 3*T) {
        break;
      }
      List<Integer> dsts = nodes.get(n);
      for (int dst : dsts) {
        if (minDist[dst] == -1 || minDist[dst] > minDist[n] + 1) {
          queue.add(dst);
          minDist[dst] = minDist[n] + 1;
        }
      }
    }
    int ans = minDist[3*T] == -1 ? -1 : minDist[3*T] / 3;
    System.out.println(ans);
  }
}
