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.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int M = scanner.nextInt();
    List<Integer>[] groups = new List[M];
    for (int i = 0; i < M; i++) groups[i] = new ArrayList<>();
    for (int i = 0; i < N; i++) {
      int a = scanner.nextInt();
      groups[a % M].add(a);
    }

    int count = groups[0].size() / 2;
    if (M % 2 == 0) count += groups[M / 2].size() / 2;
    for (int i = 1; i < M / 2 + M % 2; i++) {
      count += count(groups[i], groups[M - i]);
    }
    System.out.println(count);
  }

  private static int count(List<Integer> s, List<Integer> t) {
    if (t.size() < s.size()) return count(t, s);
    Collections.sort(t);
    int count = 0;
    for (int i = 0; i < t.size() - 1; i++) {
      if (t.get(i).equals(t.get(i + 1))) {
        count++;
        i++;
      }
    }
    return Math.min(count, (t.size() - s.size()) / 2) + s.size();
  }
}