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>
using namespace std;

int main()
{
	int N, M;
	cin >> N >> M;
	vector<int> used(1000000, 0), same(M, 0), num(M, 0);
	for (int i = 0, X; i < N; i++) {
		cin >> X;
		num[X % M]++;
		used[X]++;
		if (used[X] % 2 == 0) {
			same[X % M]++;
		}
	}
	long long res = 0;
	for (int i = 0; i < M; i++) {
		if (i == (M - i) % M) {
			res += num[i] / 2;
			num[i] %= 2;
		}
		else {
			int val = min(num[i], num[(M - i) % M]);
			if ((num[i] - val) % 2 && (num[(M - i) % M] - val) % 2 && same[i] * 2 > val && same[(M - i) % M] * 2 > val) val--;
			res += val;
			num[i] -= val;
			num[(M - i) % M] -= val;
			res += min(num[i], same[i] * 2) / 2;
			num[i] -= min(num[i], same[i] * 2) / 2 * 2;
			res += min(num[(M - i) % M], same[(M - i) % M] * 2) / 2;
			num[(M - i) % M] -= min(num[(M - i) % M], same[(M - i) % M] * 2) / 2 * 2;
		}
	}
	cout << res << endl;
	return 0;
}