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>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
//const int INF = 1e8;
using namespace std;

int main(){
	int n, m;
	cin >> n >> m;

	map<int,int> input;
	rep(i,n){
		int a;
		cin >> a;
		input[a]++;
	}
	//for(auto i : input){ cout << i.first << ' ' << i.second << endl; }

	vector<int> odd(100001,0), even(100001,0); //同値でマッチングできる:even
	for(auto i : input){
		if(i.second % 2){
			odd[i.first % m]++;
			i.second--;
		}
		even[i.first % m] += i.second;
	}


	int ans = odd[0] / 2 + even[0] / 2; //mod 0のカードのマッチング
	if(m % 2){
		range(i,1,(m + 1) / 2){
			if(odd[i] > odd[m - i]){
				swap(odd[i], odd[m - i]);
				swap(even[i], even[m - i]);
			}
			ans += odd[i];
			odd[m - i] -= odd[i]; //自分自身にマッチングできないカードを別カードにマッチング

			int p = min(odd[m - i], even[i]); //余ったカードをさらにマッチング
			ans += p;
			even[i] -= p;

			ans += even[i] / 2 + even[m - i] / 2; //自分とマッチング
		}
	}else{
		range(i,1,m / 2){
			if(odd[i] > odd[m - i]) swap(odd[i], odd[m - i]);
			ans += odd[i];
			odd[m - i] -= odd[i]; //自分自身にマッチングできないカードを別カードにマッチング

			int p = min(odd[m - i], even[i]); //余ったカードをさらにマッチング
			ans += p;
			even[i] -= p;

			ans += even[i] / 2 + even[m - i] / 2; //自分とマッチング
		}

		ans += (odd[m / 2] + even[m / 2]) / 2; // mod m / 2のカードのマッチング
	}
	cout << ans << endl;
}