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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define fbo find_by_order
#define ook order_of_key
#define INF 1e9

typedef long long ll;
typedef pair<ll,ll> ii;
typedef vector<int> vi;
typedef vector < pair<int, int> > vii;
typedef long double ld;
typedef tree<pair<int,int>, null_type, less<pair<int,int> >, rb_tree_tag, tree_order_statistics_node_update> pbds;
typedef set<int>::iterator sit;
typedef map<int,int>::iterator mit;
typedef vector<int>::iterator vit;

ll n, m, x, cnt_odd[100000], cnt[100000], ans;

int main()
{
	cin >> n >> m;
	for(int i = 0; i < n; i++){
		cin >> x;
		cnt[x%m]++;
		if(cnt[x%m]%2){
			cnt_odd[x%m]++;
		}
		else{
			cnt_odd[x%m]--;
		}
	}
	if(m%2==0){
		ans+=cnt[0]/2;
		ans+=cnt[m/2]/2;
		for(int i = 1; i < m/2; i++){
			x=min(cnt_odd[i],cnt_odd[m-i]);
			ans+=x;
			cnt_odd[i]-=x;
			cnt_odd[m-i]-=x;
			x=min(cnt_odd[i],cnt[m-i]);
			ans+=x;
			cnt_odd[i]-=x;
			cnt[m-i]-=x;
			ans+=cnt[m-i]/2;
			x=min(cnt_odd[m-i],cnt[i]);
			ans+=x;
			cnt_odd[m-i]-=x;
			cnt[i]-=x;
			ans+=cnt[i]/2;
		}
	}
	else{
		ans+=cnt[0]/2;
		for(int i = 1; i <= m/2; i++){
			x=min(cnt_odd[i],cnt_odd[m-i]);
			ans+=x;
			cnt_odd[i]-=x;
			cnt_odd[m-i]-=x;
			x=min(cnt_odd[i],cnt[m-i]);
			ans+=x;
			cnt_odd[i]-=x;
			cnt[m-i]-=x;
			ans+=cnt[m-i]/2;
			x=min(cnt_odd[m-i],cnt[i]);
			ans+=x;
			cnt_odd[m-i]-=x;
			cnt[i]-=x;
			ans+=cnt[i]/2;
		}
	}
	cout << ans;
	return 0;
}