Problem:
Score : 600 points

Problem Statement

Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.

The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.

You can change trains at a station where multiple lines are available.

The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.

Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq 2×10^5

1 \leq p_i \leq N (1 \leq i \leq M)

1 \leq q_i \leq N (1 \leq i \leq M)

1 \leq c_i \leq 10^6 (1 \leq i \leq M)

p_i \neq q_i (1 \leq i \leq M)

Input

The input is given from Standard Input in the following format:

N M
p_1 q_1 c_1
:
p_M q_M c_M

Output

Print the minimum required fare. If it is impossible to get to station N by subway, print -1 instead.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in OCaml (4.02.3) (judge verdict: Time Limit Exceeded)

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

let () = Scanf.scanf "%d %d\n" @@ fun n m ->
  let es = Array.make n [] in
  for i = 0 to m - 1 do
    Scanf.scanf "%d %d %d\n" @@ fun p q c ->
      es.(p - 1) <- (c, q - 1) :: es.(p - 1);
      es.(q - 1) <- (c, p - 1) :: es.(q - 1)
  done;
  let visited = Array.make n IntSet.empty in
  let dist = Array.make n None in
  let rec bfs01 d = function
    | [] -> ()
    | cvs ->
        List.map (fun (c, v) -> dfs c d [] v) cvs
        |> List.concat
        |> bfs01 (1 + d)
  and dfs c d acc v =
    if IntSet.mem c visited.(v) then acc
    else begin
      visited.(v) <- IntSet.add c visited.(v);
      (if dist.(v) = None then dist.(v) <- Some d);
      List.fold_left (fun acc (c', u) ->
        if c = c' then dfs c d acc u
        else (c', u) :: acc) acc es.(v)
    end in
  dist.(0) <- Some 0;
  bfs01 1 es.(0);
  match dist.(n - 1) with
  | None -> print_endline "-1"
  | Some d -> Printf.printf "%d\n" d