Problem:
Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

5 5
1 3
4 5
2 3
2 4
1 4

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.

Solution in OCaml (4.02.3) (judge verdict: Wrong Answer)

let read_ns () = Scanf.scanf "%d %d " (fun x y -> (x, y))

let bfs g goal =
  let que = Queue.create () in
  Queue.push 0 que;
  let rec doit depth reach_p =
    if reach_p && (depth = 2 || depth = 0) then "POSSIBLE"
    else if Queue.is_empty que || depth > 2 then "IMPOSSIBLE"
    else begin
      let n = Queue.pop que in
      let p = ref false in
      List.iter (fun e ->
        Queue.push e que; if e = goal then p := true)
        g.(n);
      doit (depth + 1) !p
    end in
  doit 0 false

let () =
  let (n, m) = read_ns () in
  let g = Array.make n [] in
  for _ = 0 to m - 1 do
    let (a, b) = read_ns () in
    let (a, b) = (a - 1, b - 1) in
    g.(a) <- b :: g.(a);
    g.(b) <- a :: g.(b);
  done;
  bfs g (n - 1) |> print_endline