Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2

Solution in OCaml (4.02.3) (judge verdict: Accepted)

module PQ = Set.Make(struct type t = int * int * int let compare = compare end)
let () =
  Scanf.scanf "%d %d" @@ fun n m ->
  let g = Array.init n (fun _ -> []) in
  for _ = 1 to m do
    Scanf.scanf " %d %d" @@ fun u v ->
    g.(u-1) <- v-1 :: g.(u-1);
  done;
  Scanf.scanf " %d %d" @@ fun s t ->
  let s, t = s-1, t-1 in
  let dist = Array.init 3 (fun _ -> Array.make n (max_int / 3)) in
  let rec loop pq =
    if PQ.is_empty pq then () else
    let d,x,m = PQ.min_elt pq in
    let pq = PQ.remove (d,x,m) pq in
    if d > dist.(m).(x) then loop pq else begin
      let m = (m+1) mod 3 in
      List.fold_left (fun pq y -> 
        if d+1 >= dist.(m).(y) then pq else begin
          dist.(m).(y) <- d+1;
          PQ.add (d+1, y, m) pq
        end
      ) pq g.(x) |> loop
    end
  in
  loop PQ.(singleton (0, s, 0));
  if dist.(0).(t) < max_int/3 then Printf.printf "%d\n" (dist.(0).(t) / 3)
  else print_endline "-1"