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: Wrong Answer)

#include <stdio.h>
#define MIN(x,y) (x)>(y)?(y):(x)

int main(void) {
	int n,m,i,x;
	int cnt[100000] = {0},modcnt[10000] = {0};
	
	scanf("%d %d",&n,&m);

	for (i = 0;i < n;i++) {
		scanf("%d",&x);
		cnt[x-1]++;
		modcnt[x%m]++;
	}
	
	int ans = 0;
	int canpair = 0;
	for (i = 1;i <= (m-1)/2;i++) {
		canpair = MIN(modcnt[i],modcnt[m-i]);
		ans += canpair;
		modcnt[i] -= canpair;
		modcnt[m-i] -= canpair;
	}

	if (!(m%2)) {
		canpair = modcnt[m/2] / 2;
		ans += canpair;
		modcnt[m/2] -= canpair * 2;
	}
	ans += (modcnt[0]) / 2;
	modcnt[0] -= modcnt[0] /= 2;
	
	int tmp = 0;
	for (i = 1;i < m;i++) {
		if (modcnt[i] >= 2) {
			for (tmp = i;tmp <= n;tmp+=m) {
				if (cnt[tmp-1] >= 2) {
					canpair = MIN(cnt[tmp-1] / 2,modcnt[i] / 2);
					ans += canpair;
					cnt[tmp-1] -= canpair * 2;
					modcnt[i] -= canpair * 2;
				}
				if (modcnt[i] < 2) {
					break;
				}
			}
		}
	}
	
	printf("%d\n",ans);

	return 0;
}