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: Wrong Answer)

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int m = Integer.parseInt(sa[1]);
		sa = br.readLine().split(" ");
		int[] x = new int[n];
		for (int i = 0; i < n; i++) {
			x[i] = Integer.parseInt(sa[i]);
		}
		br.close();

		int[] cntm = new int[m];
		int[] cntx = new int[100001];
		for (int i = 0; i < n; i++) {
			cntm[x[i] % m]++;
			cntx[x[i]]++;
		}

		int ans = cntm[0] / 2;
		int m2 = (m + 1) / 2;
		if (m % 2 == 0) {
			ans += cntm[m2] / 2;
		}
		for (int i = 1; i < m2; i++) {
			if (cntm[i] < cntm[m - i]) {
				ans += cntm[i];
				int d = cntm[m - i] - cntm[i];
				d /= 2;
				int same = 0;
				for (int j = i; j < cntx.length; j+=m) {
					same += cntx[j] / 2;
				}
				ans += Math.min(same, d);
			} else {
				ans += cntm[m - i];
				int d = cntm[i] - cntm[m - i];
				d /= 2;
				int same = 0;
				for (int j = m - i; j < cntx.length; j+=m) {
					same += cntx[j] / 2;
				}
				ans += Math.min(same, d);
			}
		}
		System.out.println(ans);
	}
}