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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
  static Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {
    int N = sc.nextInt();
    int M = sc.nextInt();
    ArrayList<ArrayList<Integer>> list = new ArrayList<>();
    for (int i = 0; i < M; i++) {
      list.add(new ArrayList<>());
    }
    for (int i = 0; i < N; i++) {
      int X = Integer.parseInt(sc.next());
      list.get(X % M).add(X);
    }
    int ans = list.get(0).size() / 2;
    if (M % 2 == 0) ans += list.get(M / 2).size() / 2;
    for (int i = 1; i < (M + 1) / 2; i++) {
      ArrayList<Integer> l1 = list.get(i);
      ArrayList<Integer> l2 = list.get(M - i);
      ans += Math.min(l1.size(), l2.size());
      ArrayList<Integer> l = l1.size() < l2.size() ? l2 : l1;
      HashSet<Integer> exist = new HashSet<>();
      int dup = 0;
      for (int v : l) {
        if (exist.contains(v)) {
          ++dup;
          exist.remove(v);
        } else {
          exist.add(v);
        }
      }
      ans += Math.min(dup, Math.abs(l1.size() - l2.size()) / 2);
    }
    System.out.println(ans);
  }
}