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: Time Limit Exceeded)

module S = Set.Make (struct type t = int let compare = compare end)

let () =
    let main () =
        let n, m = Scanf.sscanf (read_line ()) "%d %d" (fun n m -> n, m) in
        let uv = Array.init m (fun _ -> Scanf.sscanf (read_line ()) "%d %d" (fun u v -> u, v)) in
        let s, t = Scanf.sscanf (read_line ()) "%d %d" (fun s t -> s, t) in
        let u = Array.make (n + 1) S.empty in
        Array.iter (fun (uu, vv) -> u.(uu) <- S.add vv u.(uu)) uv;
        let v = Array.make (n + 1) S.empty in
        for i = 1 to n do
            let set = S.fold (fun k s2 -> S.union u.(k) s2) u.(i) S.empty in
            v.(i) <-  S.fold (fun k s2 -> S.union u.(k) s2) set   S.empty
        done;
        let visited = Array.make (n + 1) (-1) in
        visited.(s) <- 0;
        let rec loop step frontier =
            if S.cardinal frontier = 0 then (-1) else
                let frontier =
                    S.fold (fun e next_frontier ->
                        S.fold (fun f next_frontier ->
                            if visited.(f) >= 0 then next_frontier else (
                                visited.(f) <- step;
                                S.add f next_frontier
                            )
                        ) v.(e) next_frontier
                    ) frontier S.empty
                in
                if visited.(t) >= 0 then visited.(t) else loop (step + 1) frontier
        in
        let result = loop 1 (S.singleton s) in
        Printf.printf "%d\n" result
    in
    main ()