Problem:
Score : 700 points

Problem Statement

Takahashi is playing with N cards.

The i-th card has an integer X_i on it.

Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:

The integers on the two cards are the same.

The sum of the integers on the two cards is a multiple of M.

Find the maximum number of pairs that can be created.

Note that a card cannot be used in more than one pair.

Constraints

2≦N≦10^5

1≦M≦10^5

1≦X_i≦10^5

Input

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

N M
X_1 X_2 ... X_N

Output

Print the maximum number of pairs that can be created.

Sample Input 1

7 5
3 1 4 1 5 9 2

Sample Output 1

3

Three pairs (3,2), (1,4) and (1,9) can be created.

It is possible to create pairs (3,2) and (1,1), but the number of pairs is not maximized with this.

Sample Input 2

15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99

Sample Output 2

6

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

let () = Scanf.scanf "%d %d\n" @@ fun n m ->
  let xs = Array.init n @@ fun _ -> Scanf.scanf "%d " @@ fun x -> x in
  let count = Array.make m 0 in
  let pairs = Array.make m 0 in
  begin
    let elt = Array.make 100001 false in
    Array.iter (fun x ->
      (if elt.(x) then pairs.(x mod m) <- 1 + pairs.(x mod m));
      elt.(x) <- not elt.(x);
      count.(x mod m) <- 1 + count.(x mod m)) xs
  end;
  Printf.printf "%d\n" @@
  Array.fold_left ( + ) 0 @@ 
  Array.init m @@ fun i ->
    if m - i < i then 0
    else if i = 0 || m - i = i then count.(i) / 2
    else begin
      let x = min count.(i) count.(m - i) in
      x
      + min (count.(i) - x) (2 * pairs.(i)) / 2
      + min (count.(m - i) - x) (2 * pairs.(m - i)) / 2
    end