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>
#include<stdlib.h>

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

int cmp(const void *a,const void *b){
  return *(int *)a-*(int *)b;
}

void run(void){
  int n,m;
  scanf("%d%d",&n,&m);
  int *cnt=(int *)calloc(m,sizeof(int));
  int *x=(int *)calloc(n,sizeof(int));
  int i;
  for(i=0;i<n;i++){
    scanf("%d",x+i);
    cnt[x[i]%m]++;
  }
  qsort(x,n,sizeof(int),cmp);
  int ans=cnt[0]/2;
  cnt[0]%=2;
  for(i=1;i<m-i;i++){
    int v=MIN(cnt[i],cnt[m-i]);
    ans+=v;
    cnt[i]-=v;
    cnt[m-i]-=v;
  }
  if(i==m-i){
    ans+=cnt[i]/2;
    cnt[i]%=2;
  }
  i=0;
  while(i+1<n){
    if(x[i]==x[i+1] && cnt[x[i]%m]>=2){
      ans++;
      cnt[x[i]%m]-=2;
      i+=2;
    } else {
      i++;
    }
  }
  printf("%d\n",ans);
}

int main(void){
  run();
  return 0;
}