Problem:
Score : 700 points

Problem Statement

Aoki loves numerical sequences and trees.

One day, Takahashi gave him an integer sequence of length N, a_1, a_2, ..., a_N, which made him want to construct a tree.

Aoki wants to construct a tree with N vertices numbered 1 through N, such that for each i = 1,2,...,N, the distance between vertex i and the farthest vertex from it is a_i, assuming that the length of each edge is 1.

Determine whether such a tree exists.

Constraints

2 ≦ N ≦ 100

1 ≦ a_i ≦ N-1

Input

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

N
a_1 a_2 ... a_N

Output

If there exists a tree that satisfies the condition, print Possible. Otherwise, print Impossible.

Sample Input 1

5
3 2 2 3 3

Sample Output 1

Possible

The diagram above shows an example of a tree that satisfies the conditions. The red arrows show paths from each vertex to the farthest vertex from it.

Sample Input 2

3
1 1 2

Sample Output 2

Impossible

Sample Input 3

10
1 2 2 2 2 2 2 2 2 2

Sample Output 3

Possible

Sample Input 4

10
1 1 2 2 2 2 2 2 2 2

Sample Output 4

Impossible

Sample Input 5

6
1 1 1 1 1 5

Sample Output 5

Impossible

Sample Input 6

5
4 3 2 3 4

Sample Output 6

Possible

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

let () = Scanf.scanf "%d\n" @@ fun n ->
  let as_ = Array.init n @@ fun _ -> Scanf.scanf "%d " @@ fun a -> a in
  let d = Array.fold_left max min_int as_ in
  let count_as = Array.make n 0 in
  Array.iter (fun a -> count_as.(a) <- 1 + count_as.(a)) as_;
  print_endline @@
  if
    Array.fold_right (fun n -> ( && ) (n = 0))
      (Array.sub count_as 0 ((d + 1) / 2)) true &&
    count_as.((d + 1) / 2) = 1 + d mod 2 &&
    Array.fold_right (fun n -> ( && ) (1 < n))
      (Array.sub count_as ((d + 1) / 2 + 1) (d / 2)) true
  then "Possible"
  else "Impossible"