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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

#define min(a, b) ((a) < (b) ? a : b)

int cnt[100000];
int cnt2[100001];

int main(void) {
	int i;
	int n, m;
	int ans;

	scanf("%d %d", &n, &m);
	for (i = 0; i < n; i++) {
		int x;
		scanf("%d", &x);
		cnt[x % m]++;
		cnt2[x]++;
	}

	ans = 0;
	for (i = 1; i <= 100000; i++) {
		const int me = i % m, oppo = (m - me) % m;
		if (oppo != me)
			if (cnt[me] > cnt[oppo]) {
				const int down = min(cnt2[i] / 2, (cnt[me] - cnt[oppo]) / 2);
				cnt[i % m] -= down * 2;
				ans += down;
			}
	}

	for (i = 0; i < m; i++) {
		if ((m - i) % m == i) {
			ans += cnt[i] / 2;
			cnt[i] %= 2;
		} else {
			const int common = min(cnt[i], cnt[m - i]);
			ans += common;
			cnt[i] -= common;
			cnt[m - i] -= common;
		}
	}

	printf("%d\n", ans);

	return 0;
}